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If from any point $\mathrm{P\left(x_1, y_1\right)}$ on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}$ , tangents are drawn to the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,}$ then the corresponding chord of contact touches the another branch of the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}$ at the point
Option: 1
$\left(x_1,-y_1\right)$

Option: 2
$\left(-x_1, y_1\right)$

Option: 3
$\left(x_1, y_1\right)$

Option: 4
$\left(-x_1,-y_1\right)$

Answers (1)

best_answer

Chord of contact of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ w.r.t. point $P\left(x_1, y_1\right)$ is is

$\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1$ $...(1)$

Eq. (1) can be written as $\frac{x\left(-x_1\right)}{a^2}-\frac{y\left(-y_1\right)}{b^2}=-1$ ,

which is tangent to the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ at point $\left(-x_1,-y_1\right).$

Obviously, point $\left(x_1, y_1\right)$ and $\left(-x_1,-y_1\right)$ lie on the different branches of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1.$

Posted by

Ritika Harsh

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