The equation of the normal to the hyperbola x^2 16 - y^2 9 = 1 at… - Vidyadip

MCQ

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

✓
$y = 0$
B
$y = x$
C
$x = 0$
D
$x = - y$

✓

Answer

Correct option: A.

$y = 0$

a
(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

==> $(y - 0)\, = 0(x + 4)$

==> $y = 0.$

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