The locus of the point of intersection of the tangents to the hyperbola x^2/a^2 - y^2/b^2 = 1 which make an angle alpha with each other is:

D. (x^2+y^2-a^2+b^2)^2 = 4(x^2/a^2 - y^2/b^2 - 1) ^2 — The locus of the intersection of tangents at a fixed angle is derived from the slope condition.

A. (x^2-y^2-a^2+b^2)^2 = 4(x^2/a^2 - y^2/b^2 - 1) ^2

B. (x^2+y^2-a^2+b^2)^2 = 4(x^2/a^2 - y^2/b^2) ^2

C. (x^2+y^2-a^2+b^2)^2 = 4(x^2/a^2 - y^2/b^2 - 1) ^2

JEE Advanced · MathematicsHard

$The locus of the point of intersection of the tangents to the hyperbola \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1 which make an angle α with each other is:$

A. $(x^{2} - y^{2} - a^{2} + b^{2})^{2} = 4 (x^{2} / a^{2} - y^{2} / b^{2} - 1) tan^{2} α$
B. $(x^{2} + y^{2} - a^{2} + b^{2})^{2} = 4 (x^{2} / a^{2} - y^{2} / b^{2}) tan^{2} α$
C. $(x^{2} + y^{2} - a^{2} + b^{2})^{2} = 4 (x^{2} / a^{2} - y^{2} / b^{2} - 1) cot^{2} α$
D. $(x^{2} + y^{2} - a^{2} + b^{2})^{2} = 4 (x^{2} / a^{2} - y^{2} / b^{2} - 1) tan^{2} α$

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Correct answer: D — $(x^{2} + y^{2} - a^{2} + b^{2})^{2} = 4 (x^{2} / a^{2} - y^{2} / b^{2} - 1) tan^{2} α$

Solution

$Let the intersection point be (h, k) .$
$The slopes m_{1}, m_{2} of the tangents satisfy m_{1} + m_{2} = 2 hk / (h^{2} - a^{2}) and m_{1} m_{2} = (k^{2} + b^{2}) / (h^{2} - a^{2}) .$
$The angle condition is ∣ (m_{1} - m_{2}) / (1 + m_{1} m_{2}) ∣ = tan α .$
$Using (m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2}, we substitute the expressions.$
$Simplifying leads to (h^{2} + k^{2} - a^{2} + b^{2})^{2} = 4 (h^{2} / a^{2} - k^{2} / b^{2} - 1) a^{2} b^{2} tan^{2} α .$
$Hence the answer is (D).$

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The locus of the point of intersection of the tangents to the hyperbola a2x2​−b2y2​=1 which make an angle α with each other is:

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$The locus of the point of intersection of the tangents to the hyperbola \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1 which make an angle α with each other is:$