The product of the perpendiculars drawn from | vedclass

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Let $(x_1, y_1)$ be any point on the hyperbola,so $\frac{x_1^2}{a^2} - \f

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$\frac{a^2b^2}{a^2 + b^2}$

Vedclass · Answer

(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Let $(x_1, y_1)$ be any point on the hyperbola,so $\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$,which implies $b^2x_1^2 - a^2y_1^2 = a^2b^2$.The equations of the asymptotes are $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.The perpendicular distance from $(x_1, y_1)$ to the line $\frac{x}{a} - \frac{y}{b} = 0$ is $p_1 = \frac{|\frac{x_1}{a} - \frac{y_1}{b}|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \frac{|bx_1 - ay_1|}{ab \sqrt{\frac{a^2 + b^2}{a^2b^2}}} = \frac{|bx_1 - ay_1|}{\sqrt{a^2 + b^2}}$.The perpendicular distance from $(x_1, y_1)$ to the line $\frac{x}{a} + \frac{y}{b} = 0$ is $p_2 = \frac{|\frac{x_1}{a} + \frac{y_1}{b}|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \frac{|bx_1 + ay_1|}{\sqrt{a^2 + b^2}}$.The product of the perpendiculars is $p_1 p_2 = \frac{|b^2x_1^2 - a^2y_1^2|}{a^2 + b^2}$.Substituting $b^2x_1^2 - a^2y_1^2 = a^2b^2$,we get $p_1 p_2 = \frac{a^2b^2}{a^2 + b^2}$.

The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is

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