Question

The length of the chord of the parabola $x^2=4y$ having equation $x-\sqrt{2}y+4\sqrt{2}=0$ is :

• A. $3\sqrt{2}$
• B. $2\sqrt{11}$
• C. $6\sqrt{3}$
• D. $8\sqrt{2}$

Answer 1

The correct option is C $6\sqrt{3}$

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the point of intersection of the chord and the parabola.
$x^2=4y\cdots \left(1\right)$
$x-\sqrt{2}y+4\sqrt{2}=0\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$x-\sqrt{2}\cdot \frac{x^2}{4}+4\sqrt{2}=0$
$\Rightarrow \sqrt{2}{x}^2-4x-16\sqrt{2}=0$
$\therefore x_1+x_2=2\sqrt{2}$, $x_1{x}_2=-16$
So, $x_1-x_2=6\sqrt{2}$

Now from equation $\left(2\right)$
$x_1-\sqrt{2}{y}_1+4\sqrt{2}=0\cdots \left(a\right)$
$x_2-\sqrt{2}{y}_2+4\sqrt{2}=0\cdots \left(b\right)$
Substract $\left(a\right)$ from $\left(b\right)$
$x_2-x_1=\sqrt{2}(y_2-y_1)$
$\Rightarrow (x_2-x_1{)}^2=2(y_2-y_1{)}^2$

So, $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$=\sqrt{(x_2-x_1{)}^2+\frac{(x_2-x_1{)}^2}{2}}$
$=\sqrt{\frac{3}{2}}\times |x_2-x_1|$
$=\sqrt{\frac{3}{2}}\times 6\sqrt{2}$
$=6\sqrt{3}$