First slide

Ellipse

Question

$Let the line y = m x and the ellipse 2 x^{2} + y^{2} = 1 intersect at a point P in the first quadrant. the area of the traingle formed by the normal to this ellipse at P with axes is (N o r m a l i s p a s \sin g t h r o u g h t h e p o i n t (- \frac{1}{3 \sqrt{2}}, 0))$

Moderate

A

$\frac{2 \sqrt{2}}{3}$
B

$\frac{\sqrt{2}}{3}$
C

$\frac{2}{3}$
D

$\frac{1}{18}$

Solution

$\begin{array}{l} Let P be (x_{1}, y_{1}) \\ Equation of normal at P is \frac{x}{2 x_{1}} - \frac{y}{y_{1}} = - \frac{1}{2} \end{array}$
$\begin{array}{l} It passes through (- \frac{1}{3 \sqrt{2}}, 0) \Rightarrow \frac{- 1}{6 \sqrt{2} x_{1}} = - \frac{1}{2} \Rightarrow x_{1} = \frac{1}{3 \sqrt{2}} \\ So y_{1} = \frac{2 \sqrt{2}}{3} (as P lies in 1^{st} quadrant) \\ The area of the triangle formed by this normal with axes is \frac{1}{2} (\frac{\sqrt{2}}{3}) (\frac{1}{3 \sqrt{2}}) \\ T h e r e f o r e t h e a r e a o f t h e t r i a n g l e i s \frac{1}{18} s q u a r e u n i t s \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App