Question

Tangent to the ellipse $\frac{x^2}{4}+y^2=1$ at the point $P(\sqrt{2},\frac{1}{\sqrt{2}})$ touches the circle $x^2+y^2=r^2$ at the point $Q.$ Then the length of $PQ$ is

• A. $\frac{1}{\sqrt{10}}$
• B. $\frac{3}{\sqrt{10}}$
• C. $\frac{7}{\sqrt{10}}$
• D. $\frac{11}{\sqrt{10}}$

Answer 1

The correct option is B $\frac{3}{\sqrt{10}}$

The equation of tangent at the point $P(\sqrt{2},\frac{1}{\sqrt{2}})$ to the ellipse $x^2+4y^2=4$ is

$\sqrt{2}x+\frac{1}{\sqrt{2}}\times 4y=4\Rightarrow x+2y=2\sqrt{2}$

$r=\frac{|0+0-2\sqrt{2}|}{\sqrt{5}}=\frac{2\sqrt{2}}{\sqrt{5}}$
Let $Q$ be $(h,k)$
$\Rightarrow h^2+k^2=\frac{8}{5}.....\left(1\right)$

Radius is perpendicular to the tangent
$\Rightarrow \frac{k}{h}\times -\frac{1}{2}=-1\Rightarrow k=2h....\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$
$\Rightarrow h=\frac{2\sqrt{2}}{5}\Rightarrow k=\frac{4\sqrt{2}}{5}$

$PQ=\frac{3}{\sqrt{10}}$