Find the equation of the normal to the hyperbola $x^{2} - 4 y^{2} = 36$ at the point $(10, 4)$

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Solution

The equation of the hyperbola is
$x^{2} - 4 y^{2} = 36$ ...(i)
$\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1$
Here, $a^{2} = 36, b^{2} = 9$
$∴$ equation of the normal to the hyperbola (i) at the point $(10, 4)$ is
$\frac{a^{2} x}{x_{1}} + \frac{b^{2} y}{y_{1}} = a^{2} + b^{2}$
$\Rightarrow \frac{36 x}{10} + \frac{9 y}{4} = 36 + 9$
$\Rightarrow \frac{36 x}{10} + \frac{9 y}{4} == 45$
$\Rightarrow$ $72 x + 45 y = 900$
$\Rightarrow$ $8 x + 5 y = 100$

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