ANALYTICAL GEOMETRY OF TWO DIMENSIONS

ব্যাখ্যা (Explanation): অক্ষদ্বয়ের স্থানান্তর (translation of axes) এর ক্ষেত্রে, মূল স্থানাঙ্ক (x, y) এবং নতুন স্থানাঙ্ক (X, Y) এর মধ্যে সম্পর্কটি হল x = X + h এবং y = Y + k। এখান থেকে নতুন স্থানাঙ্ক পাওয়া যায় X = x – h এবং Y = y – k।
For translation of axes, the relation between the original coordinates (x, y) and new coordinates (X, Y) is x = X + h and y = Y + k. From this, we derive the new coordinates as X = x – h and Y = y – k.

ব্যাখ্যা (Explanation): অক্ষদ্বয়ের আবর্তনের (rotation of axes) ক্ষেত্রে, মূল স্থানাঙ্ক (x, y) এবং নতুন স্থানাঙ্ক (X, Y) এর সম্পর্কগুলি হলো:
x = X cosθ – Y sinθ
y = X sinθ + Y cosθ
Therefore, x is equal to X cosθ – Y sinθ.

ব্যাখ্যা (Explanation): আবর্তনের পর নতুন স্থানাঙ্ক X এবং Y হলে,
X² + Y² = (x cosθ + y sinθ)² + (-x sinθ + y cosθ)²
= (x²cos²θ + y²sin²θ + 2xysinθcosθ) + (x²sin²θ + y²cos²θ – 2xysinθcosθ)
= x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ)
= x² + y²
সুতরাং, x² + y² রাশিটি অপরিবর্তনীয়। এটি মূলবিন্দু থেকে বিন্দুর দূরত্বের বর্গ, যা আবর্তনের ফলে পরিবর্তিত হয় না।
Thus, the expression x² + y² is invariant. This represents the square of the distance of the point from the origin, which does not change upon rotation.

ব্যাখ্যা (Explanation): যখন অক্ষগুলি ঘোরানো হয়, তখন সাধারণ দ্বিঘাত সমীকরণের দুটি রাশি অপরিবর্তনীয় থাকে:
1. x² এবং y² এর সহগগুলির যোগফল (a + b)।
2. h² – ab।
এই অপরিবর্তনীয় রাশিগুলি কনিকের প্রকৃতি সনাক্ত করতে সাহায্য করে।
When the axes are rotated, two quantities in the general second-degree equation remain invariant:
1. The sum of the coefficients of x² and y² (a + b).
2. The expression h² – ab.
These invariants help in identifying the nature of the conic.

ব্যাখ্যা (Explanation): একটি সাধারণ দ্বিঘাত সমীকরণ একটি অধিবৃত্ত (parabola) উপস্থাপন করে যদি নির্ণায়ক (discriminant) h² – ab = 0 হয় এবং Δ ≠ 0, যেখানে Δ = abc + 2fgh – af² – bg² – ch²।
A general equation of the second degree represents a parabola if the discriminant h² – ab = 0 and Δ ≠ 0, where Δ = abc + 2fgh – af² – bg² – ch².

ব্যাখ্যা (Explanation): প্রদত্ত সমীকরণটি হলো 4x² + 9y² – 8x + 36y + 4 = 0। এখানে a = 4, b = 9, h = 0।
নির্ণায়ক h² – ab = 0² – (4)(9) = -36।
যেহেতু h² – ab < 0, তাই সমীকরণটি একটি উপবৃত্ত (ellipse) নির্দেশ করে।
The given equation is 4x² + 9y² – 8x + 36y + 4 = 0. Here, a = 4, b = 9, h = 0.
The discriminant is h² – ab = 0² – (4)(9) = -36.
Since h² – ab < 0, the equation represents an ellipse.

ব্যাখ্যা (Explanation): একটি উপবৃত্তের ক্যানোনিকাল বা প্রামাণ্য রূপ হলো X²/A² + Y²/B² = 1, যেখানে অক্ষগুলি কনিকের প্রধান অক্ষ বরাবর থাকে এবং মূলবিন্দু কনিকের কেন্দ্রে থাকে।
The canonical or standard form of an ellipse is X²/A² + Y²/B² = 1, where the axes lie along the principal axes of the conic and the origin is at the center of the conic.

ব্যাখ্যা (Explanation): সাধারণ দ্বিঘাত সমীকরণটি একজোড়া সরলরেখা প্রকাশ করে যদি নির্ণায়ক Δ = 0 হয়। নির্ণায়কটি হলো Δ = abc + 2fgh – af² – bg² – ch²। এটিকে ম্যাট্রিক্স আকারেও লেখা যায়: det([[a, h, g], [h, b, f], [g, f, c]]) = 0।
The general equation of second degree represents a pair of straight lines if the determinant Δ = 0. The determinant is Δ = abc + 2fgh – af² – bg² – ch². This can also be written in matrix form: det([[a, h, g], [h, b, f], [g, f, c]]) = 0.

ব্যাখ্যা (Explanation): ax² + 2hxy + by² = 0 সমীকরণটি মূলবিন্দুগামী দুটি সরলরেখা y = m₁x এবং y = m₂x প্রকাশ করে। এদের মধ্যে কোণ θ হলে, tanθ = |(m₁-m₂)/(1+m₁m₂)|। এখানে, m₁ + m₂ = -2h/b এবং m₁m₂ = a/b। এই মানগুলি ব্যবহার করে আমরা পাই, tanθ = |2√(h²-ab) / (a+b)|।
The equation ax² + 2hxy + by² = 0 represents two straight lines y = m₁x and y = m₂x passing through the origin. If θ is the angle between them, then tanθ = |(m₁-m₂)/(1+m₁m₂)|. Here, m₁ + m₂ = -2h/b and m₁m₂ = a/b. Using these values, we get tanθ = |2√(h²-ab) / (a+b)|.

ব্যাখ্যা (Explanation): ax² + 2hxy + by² = 0 দ্বারা প্রদত্ত সরলরেখা জোড়ার কোণের সমদ্বিখণ্ডকগুলির যৌথ সমীকরণ হলো (x²-y²)/(a-b) = xy/h। এটি একটি প্রমিত সূত্র।
The combined equation of the bisectors of the angles between the pair of lines given by ax² + 2hxy + by² = 0 is (x²-y²)/(a-b) = xy/h. This is a standard formula.

ব্যাখ্যা (Explanation): যেকোনো কনিক S = 0 এর ক্ষেত্রে, বাইরের বিন্দু (x₁, y₁) থেকে অঙ্কিত স্পর্শ জ্যা-এর (chord of contact) সমীকরণটি হলো T = 0। উপবৃত্ত x²/a² + y²/b² – 1 = 0 এর জন্য, T হলো xx₁/a² + yy₁/b² – 1। সুতরাং, সমীকরণটি হলো xx₁/a² + yy₁/b² = 1।
For any conic S = 0, the equation of the chord of contact from an external point (x₁, y₁) is given by T = 0. For the ellipse x²/a² + y²/b² – 1 = 0, T is xx₁/a² + yy₁/b² – 1. So, the equation is xx₁/a² + yy₁/b² = 1.

ব্যাখ্যা (Explanation): এটি একটি প্রমিত সূত্র। এখানে, S হলো কনিকের সমীকরণ, S₁ হলো কনিকের সমীকরণে x এবং y এর পরিবর্তে x₁ এবং y₁ বসিয়ে প্রাপ্ত মান, এবং T হলো স্পর্শ জ্যা-এর সমীকরণ (T=0)। সুতরাং, স্পর্শক জোড়ার সমীকরণ হলো SS₁ = T²।
This is a standard formula. Here, S is the equation of the conic, S₁ is the value obtained by substituting x₁ and y₁ for x and y in the equation of the conic, and T is the expression for the chord of contact (T=0). Thus, the equation of the pair of tangents is SS₁ = T².

ব্যাখ্যা (Explanation): একটি কনিকের সাপেক্ষে একটি বিন্দুর পোলারের সমীকরণটি তার স্পর্শ জ্যা-এর সমীকরণের অনুরূপ (T=0)। পরাবৃত্ত x²/a² – y²/b² = 1 এর জন্য, পোলারের সমীকরণ হলো xx₁/a² – yy₁/b² = 1।
The equation of the polar of a point with respect to a conic is identical in form to the equation of its chord of contact (T=0). For the hyperbola x²/a² – y²/b² = 1, the equation of the polar is xx₁/a² – yy₁/b² = 1.

ব্যাখ্যা (Explanation): পোলার সমীকরণ l/r = 1 + e cosθ তে ‘e’ হলো কনিকের উৎকেন্দ্রতা (eccentricity)।
– যদি e = 0, কনিকটি একটি বৃত্ত (circle)।
– যদি 0 < e < 1, কনিকটি একটি উপবৃত্ত (ellipse)।
– যদি e = 1, কনিকটি একটি অধিবৃত্ত (parabola)।
– যদি e > 1, কনিকটি একটি পরাবৃত্ত (hyperbola)।
In the polar equation l/r = 1 + e cosθ, ‘e’ is the eccentricity of the conic.
– If e = 0, it’s a circle.
– If 0 < e < 1, it's an ellipse.
– If e = 1, it’s a parabola.
– If e > 1, it’s a hyperbola.

ব্যাখ্যা (Explanation): l/r = 1 + e cosθ সমীকরণে, ‘l’ হলো অর্ধ-নাভিলম্ব (semi-latus rectum)। সম্পূর্ণ নাভিলম্বের দৈর্ঘ্য হলো 2l।
In the equation l/r = 1 + e cosθ, ‘l’ represents the semi-latus rectum. The length of the full latus rectum is 2l.

ব্যাখ্যা (Explanation): এটি একটি প্রমিত সূত্র। l/r = 1 + e cosθ কনিকের উপর ‘α’ বিন্দুতে স্পর্শকের পোলার সমীকরণটি হলো l/r = e cosθ + cos(θ – α)।
This is a standard formula. The polar equation of the tangent to the conic l/r = 1 + e cosθ at the point with vectorial angle ‘α’ is l/r = e cosθ + cos(θ – α).

ব্যাখ্যা (Explanation): Let the origin be shifted to (h, k). New coordinates (X, Y) are given by X = x – h, Y = y – k.
For the point (3, 4), the new coordinates are (-1, 2).
So, -1 = 3 – h => h = 4.
And, 2 = 4 – k => k = 2.
The origin is shifted to (4, 2).
Now, for the point (5, 5), the new coordinates will be:
X = 5 – 4 = 1
Y = 5 – 2 = 3
The new coordinates are (1, 3).

ব্যাখ্যা (Explanation): We use the rotation formulas: x = X cosθ – Y sinθ and y = X sinθ + Y cosθ.
Substituting these into xy = c²:
(X cosθ – Y sinθ)(X sinθ + Y cosθ) = c²
X²cosθsinθ + XY(cos²θ – sin²θ) – Y²sinθcosθ = c²
(X² – Y²)sinθcosθ + XYcos(2θ) = c²
(X² – Y²)(sin(2θ)/2) + XYcos(2θ) = c²
To get the form X² – Y² = 2c², the coefficient of XY must be zero. So, cos(2θ) = 0.
This means 2θ = 90°, so θ = 45°.
If θ = 45°, sin(2θ) = sin(90°) = 1. The equation becomes (X² – Y²)(1/2) = c², which is X² – Y² = 2c².

ব্যাখ্যা (Explanation): For the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 to represent a circle, two conditions must be met:
1. The coefficient of x² must be equal to the coefficient of y² (a = b).
2. The coefficient of the xy term must be zero (h = 0).
This reduces the equation to a(x² + y²) + 2gx + 2fy + c = 0, which is the general form of a circle.

ব্যাখ্যা (Explanation): Here a=1, h=1, b=1. The condition for a parabola is h² – ab = 0.
1² – (1)(1) = 1 – 1 = 0.
Since h² – ab = 0, the conic is a parabola. A parabola has its centre at infinity.
Alternatively, trying to find the center by solving ∂S/∂x = 0 and ∂S/∂y = 0:
∂S/∂x = 2x + 2y – 4 = 0 => x + y = 2
∂S/∂y = 2x + 2y – 6 = 0 => x + y = 3
These two lines are parallel and never intersect, which implies the center is at infinity.

ব্যাখ্যা (Explanation): We factorize the homogeneous equation:
x² – 5xy + 4y² = 0
x² – xy – 4xy + 4y² = 0
x(x – y) – 4y(x – y) = 0
(x – y)(x – 4y) = 0
So, the two lines are x – y = 0 and x – 4y = 0.

ব্যাখ্যা (Explanation): The angle θ between the lines is given by tanθ = |2√(h²-ab) / (a+b)|.
For the lines to be perpendicular, θ = 90°, which means tanθ is undefined. This happens when the denominator is zero.
Therefore, a + b = 0. This is the condition for the lines to be perpendicular.
Alternatively, if the slopes are m₁ and m₂, then m₁m₂ = -1. From the equation, m₁+m₂ = -2h/b and m₁m₂ = a/b. So, a/b = -1, which implies a = -b or a + b = 0.

ব্যাখ্যা (Explanation): We homogenize the equation of the circle using the equation of the line.
The line is x + y = 1. The circle is x² + y² = 4.
We can write the circle equation as x² + y² = 4(1)².
Now substitute 1 = (x + y).
x² + y² = 4(x + y)²
x² + y² = 4(x² + 2xy + y²)
x² + y² – 4x² – 8xy – 4y² = 0
-3x² – 8xy – 3y² = 0
3x² + 8xy + 3y² = 0.

ব্যাখ্যা (Explanation): Let the pole be (x₁, y₁). The equation of the polar is xx₁/9 + yy₁/4 = 1.
This is the same line as 2x – 3y = 6. To compare them, we write the given line in the same format: (2x – 3y)/6 = 1, which is x/3 – y/2 = 1.
Comparing the coefficients of x and y from both equations:
x₁/9 = 1/3 => x₁ = 9/3 = 3.
y₁/4 = -1/2 => y₁ = -4/2 = -2.
So the pole is (3, -2).

ব্যাখ্যা (Explanation): In polar coordinates, r represents the distance from the pole (origin). The equation r = 4 means that for any angle θ, the distance from the pole is always 4. This is the definition of a circle with its center at the pole and a radius of 4.
In Cartesian coordinates, r = √(x² + y²). So, √(x² + y²) = 4, which gives x² + y² = 16. This is clearly a circle centered at the origin with radius 4.

ব্যাখ্যা (Explanation): For translation, we substitute x = X + h and y = Y + k. Here, (h, k) = (1, 2).
So, x = X + 1 and y = Y + 2.
The new equation is 3(X + 1) – 4(Y + 2) + 5 = 0.
3X + 3 – 4Y – 8 + 5 = 0.
3X – 4Y = 0.

ব্যাখ্যা (Explanation): The new coordinates (X, Y) are given by:
X = x cosθ + y sinθ
Y = -x sinθ + y cosθ
Here, (x, y) = (√2, √2) and θ = 45°.
X = √2 cos(45°) + √2 sin(45°) = √2(1/√2) + √2(1/√2) = 1 + 1 = 2.
Y = -√2 sin(45°) + √2 cos(45°) = -√2(1/√2) + √2(1/√2) = -1 + 1 = 0.
The new coordinates are (2, 0).

ব্যাখ্যা (Explanation): To remove the X² and Y² terms and introduce an XY term, we need to rotate by an angle θ such that the new coefficients of x’² and y’² become zero. The transformed equation for x²-y²=a² is (Xcosθ-Ysinθ)² – (Xsinθ+Ycosθ)² = a². The coefficient of X² is cos²θ-sin²θ = cos(2θ) and for Y² is sin²θ-cos²θ=-cos(2θ). To eliminate them, we need cos(2θ)=0, so 2θ = ±90°, which means θ = ±45°. Let’s test θ = -45°.
x = Xcos(-45) – Ysin(-45) = X(1/√2) + Y(1/√2)
y = Xsin(-45) + Ycos(-45) = -X(1/√2) + Y(1/√2)
x² – y² = [ (X+Y)/√2 ]² – [ (Y-X)/√2 ]² = (1/2)[ (X²+2XY+Y²) – (Y²-2XY+X²) ] = (1/2)[4XY] = 2XY.
So 2XY = a², which gives XY = a²/2. A rotation by -45° works.

ব্যাখ্যা (Explanation): Let S = 14x² – 4xy + 11y² – 44x – 58y + 71.
To find the centre, we solve ∂S/∂x = 0 and ∂S/∂y = 0.
∂S/∂x = 28x – 4y – 44 = 0 => 7x – y = 11 (Eq. 1)
∂S/∂y = -4x + 22y – 58 = 0 => -2x + 11y = 29 (Eq. 2)
From (1), y = 7x – 11. Substitute into (2):
-2x + 11(7x – 11) = 29
-2x + 77x – 121 = 29
75x = 150 => x = 2.
Then y = 7(2) – 11 = 14 – 11 = 3.
The centre is (2, 3).

ব্যাখ্যা (Explanation): We check the discriminant h² – ab.
Here, a = 9, b = 16, and 2h = -24 so h = -12.
h² – ab = (-12)² – (9)(16) = 144 – 144 = 0.
Since h² – ab = 0, the conic is a parabola (assuming Δ ≠ 0, which is generally the case unless it’s a pair of parallel lines).

ব্যাখ্যা (Explanation): To find the point of intersection, we solve the equations ∂S/∂x = 0 and ∂S/∂y = 0.
S = 2x² + 3xy – 2y² + x + 7y – 3.
∂S/∂x = 4x + 3y + 1 = 0 (Eq. 1)
∂S/∂y = 3x – 4y + 7 = 0 (Eq. 2)
Multiply (1) by 4 and (2) by 3:
16x + 12y + 4 = 0
9x – 12y + 21 = 0
Adding the two new equations: 25x + 25 = 0 => x = -1.
Substitute x = -1 into (1): 4(-1) + 3y + 1 = 0 => -3 + 3y = 0 => y = 1.
The point of intersection is (-1, 1).

ব্যাখ্যা (Explanation): The angle θ between the lines is given by tanθ = |2√(h²-ab) / (a+b)|.
For the lines to be coincident, the angle between them must be 0.
tan(0) = 0. This requires the numerator to be zero.
So, 2√(h²-ab) = 0, which means h² – ab = 0.

ব্যাখ্যা (Explanation): First, check h² – ab. Here a=4, h=2, b=1.
h² – ab = 2² – 4*1 = 0. This indicates a parabola or parallel lines.
Let’s try to factor the equation. The first part is (2x+y)².
The equation is (2x+y)² – 2(2x+y) – 3 = 0.
Let z = 2x+y. The equation becomes z² – 2z – 3 = 0.
Factoring this quadratic: (z – 3)(z + 1) = 0.
Substituting back z = 2x+y, we get (2x+y-3)(2x+y+1) = 0.
This represents two parallel straight lines: 2x+y-3=0 and 2x+y+1=0.

ব্যাখ্যা (Explanation): From the previous question, we found the lines to be 2x+y-3=0 and 2x+y+1=0.
The distance between two parallel lines Ax+By+C₁=0 and Ax+By+C₂=0 is given by the formula d = |C₁ – C₂| / √(A² + B²).
Here, A=2, B=1, C₁=-3, C₂=1.
d = |-3 – 1| / √(2² + 1²) = |-4| / √(4 + 1) = 4/√5.

ব্যাখ্যা (Explanation): The director circle is the locus of the point of intersection of perpendicular tangents to the ellipse.
For an ellipse x²/a² + y²/b² = 1, the equation of the director circle is x² + y² = a² + b².
Here, a² = 16 and b² = 9.
So, the equation is x² + y² = 16 + 9 = 25.

ব্যাখ্যা (Explanation): The equation of the hyperbola is x²/9 – y²/9 = 1.
Let the pole be (x₁, y₁). The equation of its polar is xx₁/9 – yy₁/9 = 1, which simplifies to xx₁ – yy₁ = 9.
The given line is x + y = -1.
Comparing the two line equations: xx₁ – yy₁ = 9 and x + y = -1.
We can write the second line as -x – y = 1.
Let’s compare coefficients: x₁/(-1) = (-y₁)/(-1) = 9/1.
From x₁/(-1) = 9, we get x₁ = -9.
From y₁/1 = 9, we get y₁ = 9.
So, the pole is (-9, 9).

ব্যাখ্যা (Explanation): Two points are said to be conjugate with respect to a conic if the polar of one point passes through the other. If the polar of P(x₁, y₁) is perpendicular to the polar of Q(x₂, y₂), then P and Q are conjugate points. This is a fundamental property of poles and polars.
দুটি বিন্দুকে একটি কনিকের সাপেক্ষে অনুবন্ধী বলা হয় যদি একটির পোলার অন্য বিন্দুগামী হয়। যদি P(x₁, y₁) বিন্দুর পোলার Q(x₂, y₂) বিন্দুর পোলারের উপর লম্ব হয়, তবে P এবং Q অনুবন্ধী বিন্দু। এটি পোল এবং পোলারের একটি মৌলিক ধর্ম।

ব্যাখ্যা (Explanation): A straight line passing through the pole (origin) maintains a constant angle with the initial axis (the positive x-axis). The radial distance ‘r’ can vary, but the vectorial angle ‘θ’ is fixed for all points on the line. Therefore, its equation is θ = constant.
মেরুবিন্দুগামী (মূলবিন্দু) একটি সরলরেখা প্রারম্ভিক অক্ষের (ধনাত্মক x-অক্ষ) সাথে একটি ধ্রুবক কোণ বজায় রাখে। ‘r’-এর মান পরিবর্তন হতে পারে, কিন্তু রেখার সমস্ত বিন্দুর জন্য ভেক্টর কোণ ‘θ’ স্থির থাকে। অতএব, এর সমীকরণ হল θ = ধ্রুবক।

ব্যাখ্যা (Explanation): To understand the curve, convert to Cartesian coordinates.
We know r = √(x²+y²) and x = r cosθ.
The equation is r = 2a cosθ. Multiply by r: r² = 2a (r cosθ).
Substitute the Cartesian equivalents: x² + y² = 2ax.
Rearranging, we get x² – 2ax + y² = 0.
Completing the square for x: (x² – 2ax + a²) – a² + y² = 0.
(x – a)² + y² = a².
This is the equation of a circle with center (a, 0) and radius ‘a’. This circle passes through the origin (pole), since (0-a)²+0² = a².

ব্যাখ্যা (Explanation): The equation of a tangent to the conic l/r = 1 + e cosθ at point ‘α’ is l/r = cos(θ-α) + e cosθ.
l/r = cosθcosα + sinθsinα + e cosθ = (cosα + e)cosθ + (sinα)sinθ.
We are given the line l/r = A cosθ + B sinθ.
Comparing the two equations, we get A = cosα + e and B = sinα.
From these, A – e = cosα and B = sinα.
Squaring and adding: (A – e)² + B² = cos²α + sin²α = 1.
This is the required condition.

ব্যাখ্যা (Explanation): After rotating the axes by an angle θ, the new coefficient of the XY term in the transformed equation is (b-a)sin(2θ) + 2hcos(2θ). To eliminate this term, we set it to zero.
(a-b)sin(2θ) = 2hcos(2θ)
tan(2θ) = 2h / (a-b)
So, the required angle of rotation is θ = (1/2) tan⁻¹(2h / (a-b)).

ব্যাখ্যা (Explanation): A hyperbola is called rectangular if its asymptotes are perpendicular. The angle between the asymptotes is the same as the angle between the lines ax² + 2hxy + by² = 0. For these lines to be perpendicular, the condition is a + b = 0. Also, for it to be a hyperbola, we need h² – ab > 0. The condition a+b=0 ensures this (unless a=b=0), because if a = -b, then h² – ab = h² – a(-a) = h² + a² which is always > 0 (for real non-zero coefficients).

ব্যাখ্যা (Explanation): For the pair of lines ax² + 2hxy + by² = 0, the condition for them to be perpendicular is a + b = 0.
In the given equation, a = 3 and b = -3.
So, a + b = 3 + (-3) = 0.
Therefore, the lines are perpendicular.

ব্যাখ্যা (Explanation): The equation of the chord of any conic S=0 with a given midpoint (x₁, y₁) is given by the formula T = S₁.
Here, S = x²/a² – y²/b² – 1.
T = xx₁/a² – yy₁/b² – 1.
S₁ = x₁²/a² – y₁²/b² – 1.
So, the equation is xx₁/a² – yy₁/b² – 1 = x₁²/a² – y₁²/b² – 1, which simplifies to xx₁/a² – yy₁/b² = x₁²/a² – y₁²/b².

ব্যাখ্যা (Explanation): We substitute x = r cosθ and y = r sinθ into the Cartesian equation.
(r cosθ – a)² + (r sinθ – b)² = c²
(r²cos²θ – 2ar cosθ + a²) + (r²sin²θ – 2br sinθ + b²) = c²
r²(cos²θ + sin²θ) – 2r(a cosθ + b sinθ) + a² + b² = c²
r² – 2r(a cosθ + b sinθ) + a² + b² – c² = 0.

ব্যাখ্যা (Explanation): When the axes are translated (origin shifted), the x and y terms are replaced by X+h and Y+k. This only affects the linear (x, y) and constant terms. The coefficients of the second-degree terms (x², xy, y²) do not change.
Thus, a, h, and b are invariants under translation.

ব্যাখ্যা (Explanation): The new origin is the center of the conic. We find it by solving ∂S/∂x = 0 and ∂S/∂y = 0.
S = 3x² + 2xy + 3y² – 4x – 4y
∂S/∂x = 6x + 2y – 4 = 0 => 3x + y = 2
∂S/∂y = 2x + 6y – 4 = 0 => x + 3y = 2
Multiplying the first equation by 3: 9x + 3y = 6.
Subtracting the second equation (x + 3y = 2) from this: 8x = 4 => x = 1/2.
Substitute x=1/2 into 3x+y=2: 3(1/2) + y = 2 => y = 2 – 3/2 = 1/2.
The center, and thus the new origin, is (1/2, 1/2).

ব্যাখ্যা (Explanation): The equation ax² + 2hxy + by² = 0 represents a pair of lines passing through the origin (0, 0).
The distance from the origin to any line passing through the origin is always zero.
Therefore, the product of the distances is 0 * 0 = 0.
Note: For the general equation ax²+2hxy+by²+2gx+2fy+c=0, the product is |c| / √((a-b)² + 4h²).

ব্যাখ্যা (Explanation): The director circle is the locus of the intersection of perpendicular tangents. For a hyperbola x²/a² – y²/b² = 1, its equation is x² + y² = a² – b².
Note that the director circle is real only if a > b. If a < b, there are no perpendicular tangents.

ব্যাখ্যা (Explanation): The definition of a conic in polar coordinates is SP = e * PM, where S is the focus (pole), P is a point (r, θ) on the conic, and PM is the perpendicular distance from P to the directrix.
For the equation l/r = 1 + e cosθ, the directrix is a vertical line. The distance from the pole to the directrix is d = l/e.
The equation of this vertical line is x = d, or r cosθ = l/e.
Rearranging gives l/r = e cosθ.

ব্যাখ্যা (Explanation): This equation has no linear terms, so only rotation is needed. The angle of rotation is θ = (1/2)tan⁻¹(2h/(a-b)) = (1/2)tan⁻¹(2/(3-3)), which is 45°.
Substitute x = (X-Y)/√2, y = (X+Y)/√2.
3((X-Y)/√2)² + 2((X-Y)/√2)((X+Y)/√2) + 3((X+Y)/√2)² = 2
3/2(X²-2XY+Y²) + 2/2(X²-Y²) + 3/2(X²+2XY+Y²) = 2
Multiply by 2: 3X²-6XY+3Y² + 2X²-2Y² + 3X²+6XY+3Y² = 4
8X² + 4Y² = 4 => 2X² + Y² = 1.
This can be written as X²/(1/2) + Y²/1 = 1. All options represent the same canonical form.

ব্যাখ্যা (Explanation): We can rearrange the equation as (x² – 2x + 1) – y² = 0.
This is (x – 1)² – y² = 0.
Using the difference of squares formula (a²-b² = (a-b)(a+b)), we get:
((x-1) – y)((x-1) + y) = 0.
So the two lines are x – y – 1 = 0 and x + y – 1 = 0.

ব্যাখ্যা (Explanation): This is a standard formula in polar coordinates. The equation of a chord joining two points with vectorial angles (α-β) and (α+β) on the conic l/r = 1 – e cosθ is given by l/r = secβ cos(θ-α) – e cosθ. The derivation involves solving a system of two equations for the line that passes through the two given points on the conic.

ব্যাখ্যা (Explanation): Let the ellipse be x²/a² + y²/b² = 1. One focus is at S(ae, 0).
The equation of the polar of (x₁, y₁) is xx₁/a² + yy₁/b² = 1.
For the focus S(ae, 0), the polar is x(ae)/a² + y(0)/b² = 1.
This simplifies to x(e/a) = 1, or x = a/e.
x = a/e is the equation of the corresponding directrix. This is a fundamental property connecting the focus, polar, and directrix.

ব্যাখ্যা (Explanation): The equation of the pair of asymptotes of a hyperbola is found by changing the constant term in its standard equation to zero.
For the hyperbola x²/a² – y²/b² = 1, the combined equation of the asymptotes is x²/a² – y²/b² = 0.
This can be factored as (x/a – y/b)(x/a + y/b) = 0, which gives the two separate asymptote lines y = (b/a)x and y = -(b/a)x.

ব্যাখ্যা (Explanation): This is an application of homogenization. We need to make the conic equation homogeneous of degree 2 using the line equation.
From the line, lx+my=n, we get (lx+my)/n = 1.
The conic is ax²+2hxy+by²=1. We can write it as ax²+2hxy+by² = (1)².
Substitute the expression for 1 from the line equation:
ax²+2hxy+by² = ((lx+my)/n)²
n²(ax²+2hxy+by²) = (lx+my)².

ব্যাখ্যা (Explanation): The equation of the normal to the parabola y²=4ax in terms of its slope ‘m’ is y = mx – 2am – am³. This equation is a cubic in ‘m’. For a given point (h, k) through which the normal passes, we get k = mh – 2am – am³, which is a cubic equation in ‘m’. A cubic equation can have at most three real roots for ‘m’. Therefore, a maximum of three normals can be drawn from a given point to a parabola.

ব্যাখ্যা (Explanation): This is the standard condition of tangency for an ellipse. To derive it, substitute y = mx + c into the ellipse equation and set the discriminant of the resulting quadratic equation in x to zero (since a tangent intersects the curve at exactly one point). The resulting condition is c² = a²m² + b².

ব্যাখ্যা (Explanation): This is the normal form of a straight line in polar coordinates. A point (r, θ) on the line forms a right-angled triangle with the pole and the foot of the perpendicular from the pole to the line. The angle between the position vector of P and the normal is (θ-α). From trigonometry, the projection of the position vector OP onto the normal is p. So, r cos(θ-α) = p.

ব্যাখ্যা (Explanation): We can try to factor the expression.
xy – 2x – y + 1 = 0
x(y-2) – (y-1) = 0. This doesn’t seem to factor directly.
Let’s try x(y-2) – y + 2 – 1 = 0 => x(y-2) – 1(y-2) – 1 = 0 => (x-1)(y-2) = 1.
This is a rectangular hyperbola with center at (1, 2) and asymptotes x=1, y=2.
Let’s re-check the question for a typo. If the question was xy – 2x – y + 2 = 0.
xy – 2x – y + 2 = x(y-2) – 1(y-2) = (x-1)(y-2) = 0. This would be a pair of lines x=1 and y=2.
Let’s check Δ for the original equation: a=0, b=0, c=1, h=1/2, g=-1, f=-1/2.
Δ = abc + 2fgh – af² – bg² – ch² = 0 + 2(-1/2)(-1)(1/2) – 0 – 0 – 1(1/2)² = 1/2 – 1/4 = 1/4 ≠ 0.
h²-ab = (1/2)² – 0*0 = 1/4 > 0. So it is a hyperbola.
a+b = 0+0 = 0. So it is a rectangular hyperbola.
Okay, the original equation is a hyperbola. Let’s assume the question intended to ask for pair of lines. Let’s create a better question.
Corrected Question: What conic does xy – 2x – 3y + 6 = 0 represent?
Answer: Pair of Lines.
Explanation: xy – 2x – 3y + 6 = x(y-2) – 3(y-2) = (x-3)(y-2) = 0. This represents the pair of lines x=3 and y=2.

ব্যাখ্যা (Explanation): Two pairs of lines are equally inclined if their angle bisectors are the same.
The equation of the bisectors for the first pair is (x²-y²)/(a-b) = xy/h.
The equation of the bisectors for the second pair is (x²-y²)/(a’-b’) = xy/h’.
For these to be the same line, the coefficients must be proportional: 1/(a-b) / (1/(a’-b’)) = 1/h / (1/h’).
This simplifies to (a’-b’)/(a-b) = h’/h, or h/(a-b) = h’/(a’-b’).

ব্যাখ্যা (Explanation): Let the pole be (h, k). The equation of its polar with respect to y² = 4ax is yk = 2a(x+h).
This is 2ax – ky + 2ah = 0.
This line is identical to the given line lx + my = 1, or lx + my – 1 = 0.
Comparing coefficients: 2a/l = -k/m = 2ah/(-1).
From the first and third parts: 2a/l = -2ah => h = -1/l.
From the second and third parts: -k/m = -2ah => k = 2amh.
Substitute h = -1/l: k = 2am(-1/l) = -2am/l.
The locus is y = -2am/l, or ly + 2am = 0, which is a straight line.

ব্যাখ্যা (Explanation): We use the formula SS₁ = T².
Here, the point is the origin (0, 0).
S = x² + y² + 2gx + 2fy + c
S₁ = S(0,0) = 0² + 0² + 2g(0) + 2f(0) + c = c
T = x(0) + y(0) + g(x+0) + f(y+0) + c = gx + fy + c
The equation is (x² + y² + 2gx + 2fy + c) * c = (gx + fy + c)²
c(x²+y²) + 2c(gx+fy) + c² = (gx+fy)² + 2c(gx+fy) + c²
c(x²+y²) = (gx+fy)². This is the required equation.

ব্যাখ্যা (Explanation): The director circle is the locus of the point of intersection of perpendicular tangents. For the conic l/r = 1 + e cosθ, this locus has the equation r²(1-e²) + 2ler cosθ – 2l² = 0. This is a standard result derived by finding the intersection of two perpendicular tangents in polar coordinates.

ব্যাখ্যা (Explanation): Let the lines be y = m₁x and y = m₂x. The combined equation is y²/x² – (m₁+m₂)y/x + m₁m₂ = 0.
Comparing with 4x² + 2kxy – 7y² = 0, or 7y² – 2kxy – 4x² = 0.
Dividing by 7x²: y²/x² – (2k/7)y/x – 4/7 = 0.
Sum of slopes, m₁ + m₂ = 2k/7.
Product of slopes, m₁m₂ = -4/7.
Given m₁ + m₂ = m₁m₂, so 2k/7 = -4/7.
This gives 2k = -4, so k = -2.

ব্যাখ্যা (Explanation): Multiply the equation by r: r² = ar sinθ + br cosθ.
Substitute Cartesian coordinates: x = r cosθ, y = r sinθ, r² = x² + y².
x² + y² = ay + bx.
x² – bx + y² – ay = 0.
Completing the square: (x – b/2)² – b²/4 + (y – a/2)² – a²/4 = 0.
(x – b/2)² + (y – a/2)² = (a²+b²)/4.
This is the equation of a circle with center (b/2, a/2) and radius √(a²+b²)/2. Since the equation is satisfied by (0,0), it passes through the pole (origin).

ব্যাখ্যা (Explanation): The general formula for a chord with a given midpoint is T = S₁.
For the parabola S = y² – 4ax = 0.
T = yy₁ – 2a(x+x₁).
S₁ = y₁² – 4ax₁.
So, the equation is yy₁ – 2a(x+x₁) = y₁² – 4ax₁.

ব্যাখ্যা (Explanation): For a pair of straight lines, we must have Δ = 0. For the lines to be parallel, we must have h² = ab. When these two conditions are combined, it can be shown that they imply the further condition af² = bg² (or ah² = hg² if f=0). This ensures the linear terms are consistent with two parallel lines.

ব্যাখ্যা (Explanation): First, convert the circle to Cartesian: x² + y² – 2ax = 0.
The equation of the chord of contact from (x₁, y₁) is xx₁ + yy₁ – a(x+x₁) = 0.
Now convert this back to polar: (r cosθ)(r₁ cosθ₁) + (r sinθ)(r₁ sinθ₁) – a(r cosθ + r₁ cosθ₁) = 0.
rr₁ (cosθ cosθ₁ + sinθ sinθ₁) – a(r cosθ + r₁ cosθ₁) = 0.
rr₁ cos(θ-θ₁) – a(r cosθ + r₁ cosθ₁) = 0.

ব্যাখ্যা (Explanation): If the x-axis is invariant, any point (x, 0) on it must transform to a point (X, 0) on the new X-axis. The transformation is X = x cosθ + y sinθ, Y = -x sinθ + y cosθ.
For a point (x, 0), Y = -x sinθ. For this to be 0 for all x, we need sinθ = 0.
This means θ = 0° or θ = 180°. A rotation of 0° is trivial (no change). A rotation of 180° maps the positive x-axis to the negative x-axis, but the line itself (the axis) remains the same. The y-axis is also mapped onto itself (but reversed). So, a 180° rotation leaves the axes as lines invariant.

ব্যাখ্যা (Explanation): For a hyperbola x²/a² – y²/b² = 1, eccentricity e = √(1 + b²/a²).
A rectangular hyperbola is one where the asymptotes are perpendicular, which implies a = b.
Substituting b = a, we get e = √(1 + a²/a²) = √(1 + 1) = √2.
The eccentricity of any rectangular hyperbola is always √2.

ব্যাখ্যা (Explanation): The angle bisectors of any two intersecting lines are always perpendicular to each other.
Mathematically, the equation of the bisectors is (x²-y²)/(a-b) = xy/h, or hx² – (a-b)xy – hy² = 0.
For this pair of lines, let A = h, B = -h. The sum of the coefficients of x² and y² is A+B = h + (-h) = 0.
Since the sum is zero, the lines (the bisectors) are perpendicular, and the angle between them is 90°.

ব্যাখ্যা (Explanation): The condition for y=Mx+C to be a tangent is C² = a²M² – b².
From the line lx+my+n=0, we have y = (-l/m)x – (n/m).
So, M = -l/m and C = -n/m.
Substitute these into the condition: (-n/m)² = a²(-l/m)² – b².
n²/m² = a²l²/m² – b².
Multiply by m²: n² = a²l² – b²m².
This is the required condition a²l² – b²m² = n².

ব্যাখ্যা (Explanation): The rotation formulas are x = X cosθ – Y sinθ and y = X sinθ + Y cosθ.
When these are substituted into the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0, the constant term ‘c’ is not multiplied by any variable or trigonometric function. It remains as it is. Therefore, ‘c’ is an invariant under rotation of axes.

ব্যাখ্যা (Explanation): The equation of the tangent at point t (at², 2at) on the parabola y²=4ax is ty = x + at².
The tangent at t₁ is t₁y = x + at₁².
The tangent at t₂ is t₂y = x + at₂².
Subtracting the two equations: (t₁-t₂)y = a(t₁²-t₂²) = a(t₁-t₂)(t₁+t₂).
This gives y = a(t₁+t₂).
Substituting y back into the first tangent equation: t₁[a(t₁+t₂)] = x + at₁².
at₁² + at₁t₂ = x + at₁².
This gives x = at₁t₂.
So the point of intersection is (at₁t₂, a(t₁+t₂)).

ব্যাখ্যা (Explanation): First, write the equation in standard form x²/a² + y²/b² = 1.
Divide by 45: 5x²/45 + 9y²/45 = 1 => x²/9 + y²/5 = 1.
Here, a² = 9 (so a = 3) and b² = 5.
Since a² > b², the major axis is along the x-axis.
The length of the latus rectum is given by the formula 2b²/a.
Length = 2(5) / 3 = 10/3.

ব্যাখ্যা (Explanation): We use the rotation formulas x = Xcosθ – Ysinθ and y = Xsinθ + Ycosθ. Let’s try θ = 45°.
x = (X-Y)/√2 and y = (X+Y)/√2.
x² – y² = [(X-Y)/√2]² – [(X+Y)/√2]² = 4
(1/2) * [(X²-2XY+Y²) – (X²+2XY+Y²)] = 4
(1/2) * [-4XY] = 4
-2XY = 4 => XY = -2.
So, a rotation of 45° works.

ব্যাখ্যা (Explanation): The given equation is r(1 – cosθ) = 4.
We can write it in the standard form l/r = 1 + e cos(θ – α).
The form is 4/r = 1 – cosθ = 1 + cos(θ – π).
Here, l=4, e=1, and the axis is rotated by π (it lies along the negative x-axis).
The equation of the directrix corresponding to the focus at the pole is l/r = e cos(θ-α).
So, 4/r = 1 * cos(θ-π) = -cosθ.
This gives r(-cosθ) = 4, or r cosθ = -4. (Which is x = -4 in Cartesian).

ব্যাখ্যা (Explanation): The equation can be written as (3x-y)² + 6(3x-y) + 8 = 0.
Let z = 3x – y. The equation becomes z² + 6z + 8 = 0.
Factoring gives (z+2)(z+4) = 0.
So, the two lines are 3x – y + 2 = 0 and 3x – y + 4 = 0.
The distance between them is d = |C₁ – C₂| / √(A² + B²) = |2 – 4| / √(3² + (-1)²)
d = |-2| / √(9 + 1) = 2 / √10.

ব্যাখ্যা (Explanation): The slope of the tangent at (x₁, y₁) is m_tan = -(b²/a²)(x₁/y₁).
The slope of the normal is m_norm = -1/m_tan = (a²/b²)(y₁/x₁).
The equation of the line passing through (x₁, y₁) with this slope is (y – y₁) = m_norm * (x – x₁).
(y – y₁) = (a²y₁ / b²x₁) * (x – x₁).
Rearranging gives (y – y₁) / (y₁/b²) = (x – x₁) / (x₁/a²). This is the standard form.

ব্যাখ্যা (Explanation): The pole of the line lx+my+n=0 is the point (x₁, y₁) such that its polar is the given line.
The polar of (x₁, y₁) is xx₁ + yy₁ + g(x+x₁) + f(y+y₁) + c = 0.
This is (x₁+g)x + (y₁+f)y + (gx₁+fy₁+c) = 0.
We are given that the pole is the origin (0, 0). So, x₁=0, y₁=0.
The polar of the origin is gx + fy + c = 0.
This must be the same line as lx + my + n = 0.
Comparing coefficients: l/g = m/f = n/c.
From l/g = n/c, we get lc = ng. From m/f = n/c, we get mc = nf.
This seems complicated. Let’s try the reverse.
The pole of the line lx+my+n=0 is the origin (0,0). Let the center of the circle be C(-g,-f). The pole P is (0,0). The equation of the polar of P(0,0) is gx+fy+c=0. This must be the same as lx+my+n=0.
For these to be the same line, l/g = m/f = n/c. This is the condition. Let’s re-examine the options. None match directly. There is another property: The pole of a line is the intersection of the polars of any two points on the line. Maybe a simpler property is needed.
Let’s use the formula for the pole. The pole of lx+my+n=0 is (-g – l(g²+f²-c)/(lg+mf-n), -f – m(g²+f²-c)/(lg+mf-n)).
Setting this to (0,0) is too complex.
Let’s re-evaluate: The polar of the origin (0,0) is gx+fy+c=0. If this is the line lx+my+n=0, then we must have l/g = m/f = n/c. Let this ratio be k. l=kg, m=kf, n=kc. What does lg+mf = n mean? (kg)g + (kf)f = kc => k(g²+f²) = kc. This means g²+f²=c, which is the condition that the origin is ON the circle.
Let’s try the condition that the pole of lx+my+n=0 lies on the origin. Let P(x1, y1) be the pole. The equation of the polar is x(x1+g) + y(y1+f) + (gx1+fy1+c) = 0. This is identical to lx+my+n=0. So, (x1+g)/l = (y1+f)/m = (gx1+fy1+c)/n = k. x1 = kl-g, y1 = km-f. gx1+fy1+c = kn g(kl-g) + f(km-f) + c = kn k(lg+mf) – g²-f²+c = kn k(lg+mf-n) = g²+f²-c. This gives the pole. We want the pole to be (0,0). So x1=0, y1=0. kl-g = 0 => k = g/l km-f = 0 => k = f/m So g/l = f/m => gm=fl. Also, from gx1+fy1+c = kn, we get g(0)+f(0)+c = kn => c = kn. Since k=g/l, c = (g/l)n => lc = ng. Since k=f/m, c = (f/m)n => mc = nf. The options are still not matching. Let me rethink the basic definition. The pole of a line is a point P such that the line is the polar of P. So we want the polar of the origin to be lx+my+n=0. The polar of the origin (0,0) is gx+fy+c=0. Therefore, the lines gx+fy+c=0 and lx+my+n=0 must be identical. This means their coefficients are proportional: g/l = f/m = c/n. From g/l=c/n, we get ng = lc. From f/m=c/n, we get nf = mc. This is the correct condition. Now let’s check the options again. (C) lg + mf = n. Let’s assume n=1. Then lx+my+1=0. The pole is (0,0). The polar of (0,0) is gx+fy+c=0. So lx+my+1=0 and gx+fy+c=0 are the same. l=g/c, m=f/c. Let’s check option C. l*g+m*f = (g/c)*g + (f/c)*f = (g²+f²)/c. We want this to be n=1. So g²+f²=c. This is the condition for origin to be ON the circle, not for its polar to be a specific line. There must be a mistake in my understanding or a common formula I’m forgetting. Let’s try another approach. The center is (-g,-f). The radius is R=sqrt(g²+f²-c). The distance from the center to the polar of P is R²/CP. The distance from the center to the line lx+my+n=0 is d = |l(-g)+m(-f)+n|/sqrt(l²+m²). The pole P is (0,0). The center is C(-g,-f). CP = sqrt(g²+f²). The polar of P(0,0) is gx+fy+c=0. The line given is lx+my+n=0. So g/l = f/m = c/n. This is the condition. Let’s re-read the question carefully. “If the origin is a pole of the line…”. This means the polar of the origin is the line. My derivation g/l=f/m=c/n is correct. Why don’t the options match? Maybe there’s a typo in the question or options. Let me check for a different property. The line is perpendicular to the line joining the center and the pole. The pole is P(0,0). The center is C(-g,-f). The line PC has slope f/g. The line lx+my+n=0 has slope -l/m. So (f/g) * (-l/m) = -1 => fl = gm. This is one of the conditions we derived. What about the distance? The product of the distances from the center to the pole and to the polar line is R². CP = sqrt(g²+f²). Distance from center C(-g,-f) to line lx+my+n=0 is d = |-lg-mf+n|/sqrt(l²+m²). So, sqrt(g²+f²) * |-lg-mf+n|/sqrt(l²+m²) = g²+f²-c. This is too complicated. Let’s go back to g/l = f/m = c/n. Let’s check option C: lg + mf = n. Using g=lk and f=mk: l(lk) + m(mk) = n => k(l²+m²) = n. And c=nk. So c = n * n / (l²+m²). So c(l²+m²) = n². This is a valid condition derived from the proportionality. But option C is not this. There must be a typo in the question or options. Let’s assume the question is: If the pole of the line lx+my+n=0 with respect to circle x²+y²=a² is (x₁,y₁), what is the relation? The polar is xx₁+yy₁=a². Comparing with lx+my=-n, we get x₁/l = y₁/m = -a²/n. So x₁=-a²l/n, y₁=-a²m/n. Let’s assume the question meant: The line lx+my+n=0 passes through the pole of the line l’x+m’y+n’=0. This is reciprocity. Okay, let’s assume there is a typo in option (C) and it should be based on the proportionality. Let’s reconsider the problem statement from a different angle. The line lx+my+n=0 is the polar of the origin (0,0). The center of the circle is (-g, -f). The line joining the center (-g, -f) and the pole (0,0) must be perpendicular to the polar line (lx+my+n=0). Slope of line joining center and pole = (-f-0)/(-g-0) = f/g. Slope of polar line = -l/m. Product of slopes must be -1 for perpendicularity: (f/g) * (-l/m) = -1 => fl = gm. This is a necessary condition. Now, none of the options reflect just this. They combine terms. Let’s check my initial reasoning again. It is the most robust. Polar of (0,0) is gx+fy+c=0. This is identical to lx+my+n=0. This implies g/l = f/m = c/n. g=lk, f=mk, c=nk for some k. Let’s check the options again with this. A) c = 0 => n=0. B) n = 0 => c=0. C) lg + mf = n => l(lk) + m(mk) = nk => k(l²+m²) = nk => l²+m²=c. This is a possible condition. D) lg + mf = -nc. Wait, what if the equation of the line is lx+my=n? Then the polar is gx+fy+c=0. Proportionality is g/l = f/m = c/(-n). Let’s re-check option C: lg+mf = n. If we assume the line passes through the center… no. Let’s assume there’s a common typo where the equation is written as lx+my=1. Polar of (0,0) is gx+fy+c=0. So gx/(-c) + fy/(-c) = 1. Comparing with lx+my=1 gives l = -g/c and m = -f/c. lg+mf = (-g/c)g + (-f/c)f = -(g²+f²)/c. This doesn’t help. Let’s trust my first derivation: g/l = f/m = c/n. This gives gl=kc, fm=kc. No, g = lk, f=mk, c=nk. This is the only mathematically sound derivation. None of the options seem to follow directly without further assumptions. Let me search for this standard result. The result seems to be that the pole of lx+my+n=0 w.r.t the circle is a point (x₁,y₁). We want (x₁,y₁)=(0,0). The coordinates of the pole are given by a formula. Let’s try to find it. The pole is the intersection of polars of two points. A less error-prone way is the one I used: the polar of the required pole is the given line. Polar of (0,0) is gx+fy+c=0. Line is lx+my+n=0. g/l = f/m = c/n. This implies ng=lc and nf=mc. This IS the condition. The options provided must be wrong or there is a typo. Let me try to work backwards from option C. If lg+mf=n, how could this be true? Let’s say the circle is x²+y²+2gx+2fy=0 (passes through origin, c=0). Then polar of (x1,y1) is xx1+yy1+g(x+x1)+f(y+y1)=0. Polar of origin (0,0) is gx+fy=0. If this is lx+my+n=0, then n=0 and g/l=f/m. This does not help. Let’s re-evaluate option C: lg+mf=n. This looks very similar to the condition for a line to pass through a point. Is it possible the question is about the pole of a line w.r.t to x²+y²=r² being the center of the other circle? No. There must be a standard formula that I’m missing. Let’s try one last time. Pole of lx+my+n=0. Let it be (x₁,y₁). Polar of (x₁,y₁) is x(x₁+g) + y(y₁+f) + (gx₁+fy₁+c) = 0. This is same as lx+my+n=0. (x₁+g)/l = (y₁+f)/m = (gx₁+fy₁+c)/n. We are given (x₁,y₁) = (0,0). So, (0+g)/l = (0+f)/m = (g(0)+f(0)+c)/n g/l = f/m = c/n. This is the condition. Let’s assume this condition is true. g=lk, f=mk, c=nk. Check C: lg+mf = l(lk) + m(mk) = k(l²+m²). We want this to equal n. So k(l²+m²) = n. This gives k = n/(l²+m²). So g = ln/(l²+m²), f=mn/(l²+m²), c=n²/(l²+m²). This is a valid condition. But the option is simply lg+mf=n. What if k=1? Then g=l, f=m, c=n. Then lg+mf = l²+m². We need l²+m²=n. This is getting nowhere. The options must be based on a different interpretation or are flawed. Let me pick the most plausible-looking formula from a textbook or online resource. Searching for “pole of a line with respect to a general circle”. The condition that the pole of lx+my+n=0 with respect to circle x²+y²+2gx+2fy+c=0 is the origin, is indeed g/l = f/m = c/n. So my derivation is correct. Let’s assume there is a typo in option C and it should be ng=lc or nf=mc. Since no such option exists, let’s reconsider the problem entirely. Maybe the circle equation is different. What if the circle is x²+y²=r²? Then the polar of the origin is undefined. This question is likely flawed. But in an MCQ test, you have to choose the “best” answer. Let’s reconsider the case where the circle passes through the origin, c=0. Then polar of origin is gx+fy=0. This must be lx+my+n=0. So n=0 and g/l=f/m. This also doesn’t match the options. I will write a new question 100 which is standard and verifiable. **New Question 100:** The equation to the pair of asymptotes of the hyperbola 3x² – 5xy – 2y² + 5x + 11y – 8 = 0 is? This involves finding λ such that the general equation + λ is a pair of lines. S = 3x² – 5xy – 2y² + 5x + 11y – 8. Asymptotes are S+k=0. 3x² – 5xy – 2y² + 5x + 11y – 8 + k = 0. Here a=3, h=-5/2, b=-2, g=5/2, f=11/2, c = k-8. Condition is Δ = abc + 2fgh – af² – bg² – ch² = 0. 3(-2)(k-8) + 2(11/2)(5/2)(-5/2) – 3(11/2)² – (-2)(5/2)² – (k-8)(-5/2)² = 0 -6k + 48 – 275/4 – 3(121/4) + 2(25/4) – (k-8)(25/4) = 0 -6k + 48 – 275/4 – 363/4 + 50/4 – (25/4)k + 200/4 = 0 Multiply by 4: -24k + 192 – 275 – 363 + 50 – 25k + 200 = 0 -49k + 192 + 200 + 50 – 275 – 363 = 0 -49k + 442 – 638 = 0 -49k – 196 = 0 => k = -196/49 = -4. So the equation of asymptotes is 3x² – 5xy – 2y² + 5x + 11y – 8 – 4 = 0. 3x² – 5xy – 2y² + 5x + 11y – 12 = 0. The question should be “The asymptotes differ from the hyperbola by a constant. What is that constant?”. Answer: -4. Or the question could be: The equation of the asymptotes of 3x²… is 3x²… + k = 0. Find k. This is a good, solid final question. I will replace the problematic Q100 with this one. Final check of the generated HTML. I will use the new question 100. I’ll also re-check my Q99, it is a standard result and is correct. Q75 is also a standard result, correct. Q98 is a good problem and my solution is correct. Q95 solution is correct. Q90 solution is correct. The flow looks good. I will now generate the final output.অবশ্যই! এখানে আপনার অনুরোধ অনুযায়ী ৬০ নং প্রশ্নের পর থেকে বাকি প্রশ্নগুলো (৬১-১০০) যোগ করা হলো। “`html