Question

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

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

Answer 1

Answer: (C)

$6\sqrt{3}$

$x^2=4y$
$x-\sqrt{2}y+4\sqrt{2}=0$

Solving together we get
$x^2=4\left(\frac{x+4\sqrt{2}}{\sqrt{2}}\right)$
$\sqrt{2}{x}^2-4x-16\sqrt{2}=0$

$x_1+x_2=2\sqrt{2}$; $x_1{x}_2=\frac{-16\sqrt{2}}{\sqrt{2}}=-16$

Similarly,
$(\sqrt{2}y-4\sqrt{2}{)}^2=4y$
$2y^2+32-16y=4y$
$2y^2-20y+32=0{<}_{y_1{y}_2=16}^{y_1+y_2=10}$

${\ell }_{AB}=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$=\sqrt{(2\sqrt{2}{)}^2+64+(10{)}^2-4\left(16\right)}$

$=\sqrt{8+64+100-64}$
$=\sqrt{108}=6\sqrt{3}$