The product of the perpendiculars drawn from | vedclass

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Question

(a) We know that equation of hyperbola is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. Let

$({x_1},\,{y_1})$ be any point on hyperbola,

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) We know that equation of hyperbola is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. Let

$({x_1},\,{y_1})$ be any point on hyperbola, $	herefore \,\frac{{x_1^2}}{{{a^2}}} - \frac{{y_1^2}}{{{b^2}}} = 1$ or

${b^2}x_1^2 - {a^2}y_1^2 = {a^2}{b^2}$.

We also know that asymptotes of given hyperbola are

$\frac{{{x^2}}}{{{a^2}}}$$ - \frac{{{y^2}}}{{{b^2}}} = 0$

$	herefore \,$ Product of $ \bot $ from $({x_1},{y_1})$ to pair of lines $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 0$ is

$\frac{{|Ax_1^2 + 2H{x_1}{y_1} + By_1^2|}}{{\sqrt {{{(A - B)}^2} + 4{H^2}} }}$; $\frac{{{b^2}x_1^2 - {a^2}y_1^2}}{{\sqrt {{{({b^2} + {a^2})}^2}} }} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}$.

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

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