Question

Consider the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$. P is on it such that the radius OP(O is the center ) makes an angle $\frac{\pi }{4}$ with the positive x-axis and $\left|OP\right|=r.$. The value of r is

• A. $\frac{\sqrt{2}}{\sqrt{3}}$
• B. $\sqrt{6}$
• C. $\frac{2\sqrt{2}}{3}$
• D. $\frac{2\sqrt{2}}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{\sqrt{2}}{\sqrt{3}}$

Equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{2}=1$

$a^2=4\Rightarrow a=2$

$b^2=2\Rightarrow b=\sqrt{2}$

let $P$ be $(x_1,y_1)$ but given $OP$ makes $\frac{\pi }{4}$ angle with $x-$axis, so, $P$ lies on the ellipse.

So, $\frac{x_1^2}{4}+\frac{x_1^2}{2}=1\Rightarrow \frac{3x_1^2}{4}=1$

$\Rightarrow x_1=\frac{2}{\sqrt{3}}$

$y_1=x_1=\frac{2}{\sqrt{3}}$

So, $P(x_1,y_1)=(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}})$

$|OP|$=$r$=$\sqrt{(\frac{2}{\sqrt{3}}{)}^2+(\frac{2}{\sqrt{3}}{)}^2}=\frac{2\sqrt{2}}{\sqrt{3}}$