The locus of the point of intersection of two tangents to the ellipse $x^{2} / a^{2} + y^{2} / b^{2} = 1$ which make complementary angles with the major axis of the ellipse is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x^{2} - y^{2} = a^{2} - b^{2}$

$2 x^{2} + y^{2} = 2 a^{2} + b^{2}$

$x^{2} + y^{2} = a^{2} + b^{2}$

$x^{2} + y^{2} = a^{2} - b^{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

If $P (x_{1}, y_{1})$ is the point of intersection of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , with slopes $m_{1}$ and $m_{2} then m_{1} m_{2} = \frac{y_{1}^{2} - b^{2}}{x_{1}^{2} - a^{2}} where m_{1} = \tan θ_{1} & \tan θ_{2}$

Since $θ_{1}, θ_{2}$ are complementary angles, we have $\tan θ_{1} \tan θ_{2} = 1$

$\Rightarrow \frac{y_{1}^{2} - b^{2}}{x_{1}^{2} - a^{2}} = 1 \Rightarrow x_{1}^{2} - y_{1}^{2} = a^{2} - b^{2} . ∴ Locus of P (x_{1}, y_{1}) is x^{2} - y^{2} = a^{2} - b^{2}$