Question

If the chord of contact of tangents from two 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 rightangles, then find $\frac{x_1{x}_2}{y_1{y}_2}$.

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

Answer 1

The correct option is A $-\frac{a^4}{b^4}$

equation of chord of contact is given by $S_1=0$
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
$\frac{x{x}_2}{a^2}+\frac{y{y}_2}{b^2}=1$
since, they are perpendicular
Therefore, $m_1{m}_2=-1$
$(-\frac{x_1{b}^2}{y_1{a}^2})(-\frac{x_2{b}^2}{y_2{a}^2})=-1$
$\frac{x_1{x}_2}{y_1{y}_2}=-\frac{a^4}{b^4}$