Equation of the normal to the hyperbola | vedclass

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Question

Equation of normal to the hyperbola at the point

$(5 \sec 	heta, 4 	an 	heta)$ is

$5 \mathrm{x} \cos 	heta+4 \mathrm{y} \cot 	heta=25+16$&n

Vedclass · Accepted Answer

$\sqrt {21} \left( {x - 2y} \right) = 41$

Vedclass · Answer

Equation of normal to the hyperbola at the point

$(5 \sec 	heta, 4 	an 	heta)$ is

$5 \mathrm{x} \cos 	heta+4 \mathrm{y} \cot 	heta=25+16$      .......$(i)$

This line is perpendicular to the line $2 x+y=1$

$	herefore $ $\mathrm{m}_{1} \mathrm{m}_{2}=-1$

$\Rightarrow\left(\frac{-5 \cos 	heta}{4 \cot 	heta}ight)(-2)=-1$

$\Rightarrow$ $\sin 	heta=-\frac{2}{5}$

$	herefore $ $\cos 	heta=\sqrt{1-\frac{4}{25}}=\mp \frac{\sqrt{21}}{5}$

and $\cot 	heta=\mp \frac{\sqrt{21}}{2}$

From Eq. $(i),$

${5 x \frac{\sqrt{21}}{5}-\frac{4 y \sqrt{21}}{2}=41} $

$\Rightarrow$ ${\sqrt{21}(x-2 y)=41}$

Equation of the normal to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ perpendicular to the line $2x + y = 1$ is

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