Equation of normal to hyperbola frac{{{x^2}}}{{16}} - frac{{{y^2}}}{9} = 1 at ( - 4,0) 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 $( - 4,\;0)$ is

A

$y = 0$

B

$y = x$

C

$x = 0$

D

$x = - y$

Solution

(a) $\frac{{{x^2}}}{{16}} – \frac{{{y^2}}}{9} = 1$

$ \Rightarrow $$\frac{{2x}}{{16}} – \frac{{2y}}{9}\frac{{dy}}{{dx}} = 0$

==> $\frac{{dy}}{{dx}} = \frac{{2x \times 9}}{{16 \times 2y}}$

$ = \frac{9}{{16}}\frac{x}{y}$

==> ${\left( {\frac{{ – dx}}{{dy}}} \right)_{( – 4,0)}}$

$= \frac{{ – 16}}{9}\frac{y}{x} = 0$

Hence, equation of normal

==> $(y – 0)\, = 0(x + 4)$

==> $y = 0.$

Standard 11

Mathematics

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