A chord is drawn through the focus of the parabola y ^ 2 = 6 x such...

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 ${y}^{2} = 6x ⇒ a = \dfrac{3}{2}$

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

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(\dfrac{3}{2}, 0\right)$, equation of the chord is$y - 0 = m\left(x - \dfrac{3}{2}\right)$</p><p>$⇒ 2mx -2 y -3 m =0$</p><p>We know that the length of perpendicular from origin to a line$ax + by + c = 0$is$\dfrac{\left|c\right|}{\sqrt{{a}^{2} + {b}^{2}}}$</p><p>Given, the distance of the focal chord$2mx -2 y -3 m =0$from the vertex \left(0, 0\right)$ is $\dfrac{\sqrt{5}}{2}$

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

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

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

Squaring both sides, we get

$⇒ 5{m}^{2} + 5 = 9{m}^{2}$

$⇒ 4{m}^{2} = 5$

$⇒ m = \pm \dfrac{\sqrt{5}}{2} .$