Question

If the chords of contact of tangents from the points $(x_1,y_1)$ and $(x_2,y_2)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are at right angles, then $\frac{x_1{x}_2}{y_1{y}_2}$ is equal to:

• A. $\frac{a^2}{b^2}$
• B. $-\frac{a^2}{b^2}$
• C. $-\frac{a^4}{b^4}$
• D. $\frac{a^4}{b^4}$

Answer 1

Answer: (C)

$-\frac{a^4}{b^4}$

Chord of contact of an ellipse at $(x_1,y_1)$ is $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

and at $(x_2,y_2)$ is $\frac{x{x}_2}{a^2}+\frac{y{y}_2}{b^2}=1$

It is given that both chords are perpendicular

$\therefore m_1{m}_2=-1$
$\frac{-b^2{x}_1}{a^2{y}_1}\times \left(\frac{-b^2}{a^2}\frac{x_2}{y_2}\right)=-1$
$\therefore \frac{x_1{x}_2}{y_1{y}_2}=\frac{-a^4}{b^4}$