The product of the lengths of perpendiculars | vedclass

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(B) The equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.The equ

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$2/3$

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(B) The equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.The equations of the asymptotes to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which implies $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the lengths of the perpendiculars from $P$ to the asymptotes $L_1: \frac{x}{a} - \frac{y}{b} = 0$ and $L_2: \frac{x}{a} + \frac{y}{b} = 0$ is given by:$P = \left| \frac{\frac{x_1}{a} - \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| 	imes \left| \frac{\frac{x_1}{a} + \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| = \frac{|\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}|}{\frac{1}{a^2} + \frac{1}{b^2}}$.Since $(x_1, y_1)$ lies on the hyperbola,$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.Thus,the product is $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{1}{2} + \frac{1}{1}} = \frac{1}{3/2} = \frac{2}{3}$.

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is:

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