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IF THE CHORDS OF CONTACT OF TANGENTS FROM TWO POINTS $\left(x_1, y_1\right)$ AND $\left(x_2, y_2\right)$ 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
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$-\frac{a^4}{b^4}$
Correct Answer
$-\frac{a^4}{b^4}$
Chords of contact are $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$ and $\frac{x x_2}{a^2}+\frac{y y_2}{b^2}=1$
Product of the slopes $=-1$
$\Rightarrow-\frac{x_1 / a^2}{y_1 / b^2} \cdot-\frac{x_2 / a^2}{y_2 / b^2}=-1 \Rightarrow \frac{x_1 x_2}{a^4} \cdot \frac{b^4}{y_1 y_2}=-1 \Rightarrow \frac{x_1 x_2}{y_1 y_2}=-\frac{a^4}{b^4}$
$-\frac{a^4}{b^4}$
Chords of contact are $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$ and $\frac{x x_2}{a^2}+\frac{y y_2}{b^2}=1$
Product of the slopes $=-1$
$\Rightarrow-\frac{x_1 / a^2}{y_1 / b^2} \cdot-\frac{x_2 / a^2}{y_2 / b^2}=-1 \Rightarrow \frac{x_1 x_2}{a^4} \cdot \frac{b^4}{y_1 y_2}=-1 \Rightarrow \frac{x_1 x_2}{y_1 y_2}=-\frac{a^4}{b^4}$
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