Product of perpendiculars drawn from any point on a hyperbola to its asymptotes is

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10-2. Parabola, Ellipse, Hyperbola

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

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

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

$\frac{{ab}}{{\sqrt a + \sqrt b }}$

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

Solution

(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, $\therefore \,\frac{{x_1^2}}{{{a^2}}} – \frac{{y_1^2}}{{{b^2}}} = 1$ 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$

$\therefore \,$ Product of $ \bot $ from $({x_1},{y_1})$ to pair of lines $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 0$ is

$\frac{{|Ax_1^2 + 2H{x_1}{y_1} + By_1^2|}}{{\sqrt {{{(A – B)}^2} + 4{H^2}} }}$; $\frac{{{b^2}x_1^2 – {a^2}y_1^2}}{{\sqrt {{{({b^2} + {a^2})}^2}} }} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}$.

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