Question

The length of the chord of the parabola $y^2=4x$ which passes through the vertex and whose slope is 0.5, is .......

Answer 1

Given Parabola: $y^2=4x............\left(1\right)$

A chord of slope $\frac{1}{2}$ will be $y=\frac{1}{2}x+c............\left(2\right)$

$\left(2\right)$ passes through Vetex $V:(0,0)=>0=\frac{1}{2}0+c$

$=>c=0$

So equation $\left(2\right)$ becomes $2y=x..................\left(3\right)$

Now length of intercept of $y=mx+c$ on $y^2=4ax$ is $=\frac{4}{m^2}\sqrt{a(1+m^2)(a-mc)}$

Now the length of intercept of $y=\frac{1}{2}x$ on $y^2=4x$ will be,

$=\frac{4}{({\frac{1}{2})}^2}\sqrt{1(1+({\frac{1}{2})}^2\left)\right(1-\frac{1}{2}\left(0\right))}$

$=\frac{16}{2}\sqrt{5}$

$=8\sqrt{5}.$