Question

The length of the chord intercepted by the ellipse $x^2+2y^2=4$ on the normal at the point $(\sqrt{2},1)$ is:

• A. $\frac{3\sqrt{6}}{5}$
• B. $\frac{2\sqrt{6}}{5}$
• C. $\frac{7\sqrt{6}}{5}$
• D. $\frac{6\sqrt{6}}{5}$

Answer 1

Answer: (D)

$\frac{6\sqrt{6}}{5}$

Equation of ellipse is $x^2+2y^2=4$ .....(1)
or $\frac{x^2}{4}+\frac{y^2}{2}=1$
Here, $a^2=4,b^2=2$
Equation of normal to ellipse at $(\sqrt{2},1)$ is
$\frac{4x}{\sqrt{2}}-\frac{2y}{1}=4-2$
or $y=\sqrt{2}x-1$ ....(2)
Solving (1) and (2) simultaneously, we get
$5x^2-4\sqrt{2}x-2=0$
$\Rightarrow x_1+x_2=\frac{4\sqrt{2}}{5},x_1{x}_2=\frac{-2}{5}$
$(x_1-x_2{)}^2=(x_1+x_2{)}^2-4x_1{x}_2=\frac{16\times 2}{25}+\frac{8}{5}=\frac{72}{25}$
Using $\left(2\right)$, $(y_1-y_2{)}^2=2(x_1-x_2{)}^2=\frac{2\times 72}{25}$
The length of the chord is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$=\sqrt{\frac{3\times 72}{25}}=\frac{6\sqrt{6}}{5}$