Question

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

• A. $\frac{a^2}{c^2}+\frac{b^2}{d^2}=1$
• B. $\frac{c^2}{a^2}+\frac{d^2}{b^2}=1$
• C. $a^2{d}^2=b^2{c}^2$
• D. $a^2{d}^2+c^2{d}^2=1$

Answer 1

Answer: (A)

$\frac{a^2}{c^2}+\frac{b^2}{d^2}=1$

The equation of tangent $PQ$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1---\left(i\right)$. Let the intersection point of tangents at $P,Q$ w.r.t $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ be $C(x_1,y_1)$ so the equation would be $\frac{x{x}_1}{c^2}+\frac{y{y}_1}{d^2}=1---\left(ii\right)$

Comparing $i$ and $ii$

$\Rightarrow \frac{\frac{\cos \theta }{a}}{\frac{x_1}{c^2}}=\frac{\frac{\sin \theta }{b}}{\frac{y_1}{d^2}}=1$

$\cos \theta =\frac{a{x}_1}{c^2},\sin \theta =\frac{b{y}_1}{d^2}$

$\frac{a^2{x}_1^2}{c^4}+\frac{{b^{2}y}_1^2}{d^4}=1$

$C$ must lie on the director circle of $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$

$\Rightarrow \frac{x^2}{c^2+d^2}+\frac{y^2}{c^2+d^2}=1$

$\Rightarrow \frac{a^2}{c^4}=\frac{1}{c^2+d^2}\&\frac{b^2}{d^4}=\frac{1}{c^2+d^2}$

$\Rightarrow \frac{a^2}{c^2}+\frac{b^2}{d^2}=1$