JEE Main 2014 — Parabola Question with Solution

The focus and vertex of a parabola $y^2=4ax$ are respectively $\left(a, 0\right)$ and $\left(0, 0\right).$

The given parabola is $y^2=6x \Rightarrow a=\frac{3}{2}$

Thus, the focus and vertex are respectively $S\equiv \left(\frac{3}{2}, 0\right)$ and $V\equiv \left(0, 0\right).$

The equation of a line passing through a point $\left(x_1, y_1\right)$ and having slope $m$ is $y-y_1=m\left(x-x_1\right)$

If $m$ is the slope of the chord through focus $\left(\frac{3}{2}, 0\right)$, equation of the chord is $y-0=m\left(x-\frac{3}{2}\right)$

$\Rightarrow 2mx-2y-3m=0$

We know that the length of perpendicular from origin to a line $ax+by+c=0$ is $\frac{\left|c\right|}{\sqrt{a^2+b^2}}$

Given, the distance of the focal chord $2mx-2y-3m=0$ from the vertex $\left(0, 0\right)$ is $\frac{\sqrt{5}}{2}$

$\Rightarrow \frac{\left|-3m\right|}{\sqrt{4m^2+4}}=\frac{\sqrt{5}}{2}$

$\Rightarrow \frac{\left|-3m\right|}{2\sqrt{m^2+1}}=\frac{\sqrt{5}}{2}$

$\Rightarrow \frac{\left|-3m\right|}{\sqrt{m^2+1}}=\sqrt{5}$

Squaring both sides, we get

$\Rightarrow 5m^2+5=9m^2$

$\Rightarrow 4m^2=5$

$\Rightarrow m=\pm \frac{\sqrt{5}}{2}.$