Equation of normal to hyperbola frac{{{x^2}}}{{16}} - frac{{{y^2}}}{9} = 1 at point (8,3√ 3 ) is

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10-2. Parabola, Ellipse, Hyperbola

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The equation of the normal to the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ at the point $(8,\;3\sqrt 3 )$ is

A

$\sqrt 3 x + 2y = 25$

B

$x + y = 25$

C

$y + 2x = 25$

D

$2x + \sqrt 3 y = 25$

Solution

(d) Applying the formula, the required normal is

$\frac{{16x}}{8} + \frac{{9y}}{{3\sqrt 3 }} = 16 + 9\,\,$

$i.e.,$$2x + \sqrt 3 y = 25$

Trick : This is the only equation among the given options at which the point $(8,\,3\sqrt 3 )$ is located.

Standard 11

Mathematics

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