Question

A normal inclined at an angle of $\frac{\pi }{4}$ to the x-axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is drawn. It meets the major and minor axes in P and Q respectively. If C is the centre of the ellipse then the area of the triangle CPQ is

• A. $\frac{(a^2-b^2{)}^2}{4(a^2+b^2)}$
• B. $\frac{(a^2-b^2{)}^2}{(a^2-b^2)}$
• C. $\frac{(a^2-b^2{)}^2}{2(a^2+b^2)}$
• D. $\frac{(a^2+b^2{)}^2}{2(a^2+b^2)}$

Answer 1

The correct option is C $\frac{(a^2-b^2{)}^2}{2(a^2+b^2)}$

Given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$m={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{-b^2{x}_1}{a^2{y}_1}=-1$ ( since the normal makes an angle $\frac{\pi }{4}$ with x-axis) $\Rightarrow b^2{x}_1=a^2{y}_1\to \left(1\right)$
Since $P(x_1,y_1)$ lies the ellipse, we have $b^2{x}_1^2+a^2{y}_1^2=a^2{b}^2\to \left(2\right)$
From (1) & (2) $P=(\frac{a^2}{\sqrt{a^2+b^2}},\frac{b^2}{\sqrt{a^2+b^2}})$
$\therefore$ Equation of the normal at P is x – y = $\frac{a^2-b^2}{\sqrt{a^2+b^2}}$ $\to \left(3\right)$
Area of the $△CPQ$ ie the area of the triangle made by the line (3) with the axes is $\frac{(a^2-b^2{)}^2}{2(a^2+b^2)}$