Question

The length of the chord of the parabola $y^2=x$ which is bisected at the point $(2,1)$ is

• A. $5\sqrt{2}$
• B. $4\sqrt{5}$
• C. $4\sqrt{50}$
• D. $2\sqrt{5}$

Answer 1

Answer: (D)

$2\sqrt{5}$

Equation of chord to the given parabola with given mid point $(2,1)$ is given by,
$T=S_1$
$\Rightarrow y\left(1\right)-\frac{1}{2}(x+2)=(1{)}^2-(2)$
$\Rightarrow y=\frac{1}{2}x$
Clearly $c=0,m=\frac{1}{2}$ and $a=\frac{1}{4}$
Thus the length of the chord is $=\frac{4}{m^2}\sqrt{a(1+m^2)(a-mc)}=2\sqrt{5}$