- Home
- Standard 11
- Mathematics
10-2. Parabola, Ellipse, Hyperbola
normal
On the ellipse $\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{8} = 1$ the point $M$ nearest to the line $2x - 3y + 25 = 0$ is
A
$(-3,2)$
B
$\left( { - \sqrt 2 ,\frac{8}{3}} \right)$
C
$(3,2)$
D
$\left( {3\sqrt 2 ,0} \right)$
Solution
Let $\mathrm{M}$ be $(\sqrt{18} \cos \theta, \sqrt{8} \sin \theta) .$ As shortest
distanoe lies along common normal
Given line will be parallel to the tangent at $\mathrm{M}$
$\frac{x \cos \theta}{\sqrt{18}}+\frac{y \sin \theta}{\sqrt{8}}=1$ is parallel to
$2 x-3 y+25=0$
$\Rightarrow-\frac{\sqrt{8} \cos \theta}{\sqrt{18} \sin \theta}=\frac{2}{3} $
$\Rightarrow \tan \theta=-1$
$\Rightarrow \theta=\frac{3 \pi}{4} \Rightarrow$ point $M$ is $(-3,2)$
Standard 11
Mathematics
