- Home
- Standard 11
- Mathematics
Equation of the normal to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ perpendicular to the line $2x + y = 1$ is
$\sqrt {21} \left( {x - 2y} \right) = 41$
$x - 2y =1$
$\sqrt {41} \left( {x - 2y} \right) = 41$
$\sqrt {21} \left( {x - 2y} \right) = 21$
Solution
Equation of normal to the hyperbola at the point
$(5 \sec \theta, 4 \tan \theta)$ is
$5 \mathrm{x} \cos \theta+4 \mathrm{y} \cot \theta=25+16$ …….$(i)$
This line is perpendicular to the line $2 x+y=1$
$\therefore $ $\mathrm{m}_{1} \mathrm{m}_{2}=-1$
$\Rightarrow\left(\frac{-5 \cos \theta}{4 \cot \theta}\right)(-2)=-1$
$\Rightarrow$ $\sin \theta=-\frac{2}{5}$
$\therefore $ $\cos \theta=\sqrt{1-\frac{4}{25}}=\mp \frac{\sqrt{21}}{5}$
and $\cot \theta=\mp \frac{\sqrt{21}}{2}$
From Eq. $(i),$
${5 x \frac{\sqrt{21}}{5}-\frac{4 y \sqrt{21}}{2}=41} $
$\Rightarrow$ ${\sqrt{21}(x-2 y)=41}$
