The product of the lengths of perpendiculars | vedclass

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(B) The given 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 $b^2 = 1$.

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$2/3$

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(B) The given 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 are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which simplifies to $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$. For any point $P(x_1, y_1)$ on the hyperbola,the product of the lengths of the perpendiculars to the asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$. Substituting the values $a^2 = 2$ and $b^2 = 1$,we get: Product $= \frac{2 	imes 1}{2 + 1} = \frac{2}{3}$.

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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