Get Answers to all your Questions

header-bg qa

A tangent is drawn to \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} to cut \mathrm{\frac{x^2}{c^2}+\frac{y^2}{d^2}=1} at points P and Q. If tangent at P and Q to the ellipse \mathrm{\frac{\mathrm{x}^2}{\mathrm{c}^2}+\frac{\mathrm{y}^2}{\mathrm{~d}^2}=1} intersect at right angle, then \mathrm{\frac{a^2}{c^2}+\frac{b^2}{d^2}=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let tangents to \mathrm{\frac{x^2}{c^2}+\frac{y^2}{d^2}=1} at P and Q intersect at (h, k) at right angle

\mathrm{\therefore \mathrm{h}^2+\mathrm{k}^2=\mathrm{c}^2+\mathrm{d}^2 \quad} (director circle)                     .....(1)

and chord of contact PQ is given by

\mathrm{ \frac{h x}{c^2}+\frac{k y}{d^2}=1 }                                                             .....(2)

which is tangent to \mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1}

Equation of tangent to \mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1} is

\mathrm{ \frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1 }
Equation (2) and (3) represent the same line

\mathrm{ \begin{aligned} & \therefore \frac{h a}{c^2 \cos \theta}=\frac{k b}{d^2 \sin \theta}=1 \\\\ & \therefore \quad \frac{h^2 a^2}{c^4}+\frac{k^2 b^2}{d^4}=1 \end{aligned} }
Comparing with equation (1), \mathrm{\frac{a^2}{c^4}=\frac{b^2}{d^4}=\frac{1}{c^2+d^2} \Rightarrow \frac{a^2}{c^2}=\frac{c^2}{c^2+d^2} \quad \& \frac{b^2}{d^2}=\frac{d^2}{c^2+d^2}}

Hence \mathrm{\frac{a^2}{c^2}+\frac{b^2}{d^2}=\frac{c^2+d^2}{c^2+d^2}=1.}

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE