JEE Main 2014 — Mathematics Coordinate Geometry

The focus and vertex of a parabola ${y}^{2}=4ax$y2=4ax are respectively $(a,0)$(a,0) and $(0,0).$(0,0).

The given parabola is ${y}^{2}=6x\Rightarrow a=\frac{3}{2}$y2=6x⇒a=23​

Thus, the focus and vertex are respectively $S\equiv (\frac{3}{2},0)$S≡(23​,0) and $V\equiv (0,0).$V≡(0,0).

The equation of a line passing through a point $({x}_{1},{y}_{1})$(x1​,y1​) and having slope $m$m is $y-{y}_{1}=m(x-{x}_{1})$y−y1​=m(x−x1​)

If $m$m is the slope of the chord through focus $(\frac{3}{2},0)$(23​,0), equation of the chord is $y-0=m(x-\frac{3}{2})$y−0=m(x−23​)

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

We know that the length of perpendicular from origin to a line $ax+by+c=0$ax+by+c=0 is $\frac{|c|}{\sqrt{{a}^{2}+{b}^{2}}}$a2+b2​∣c∣​

Given, the distance of the focal chord $2mx-2y-3m=0$2mx−2y−3m=0 from the vertex $(0,0)$(0,0) is $\frac{\sqrt{5}}{2}$25​​

$$\Rightarrow \frac{|-3m|}{\sqrt{4{m}^{2}+4}}=\frac{\sqrt{5}}{2}$$⇒4m2+4​∣−3m∣​=25​​

$$\Rightarrow \frac{|-3m|}{2\sqrt{{m}^{2}+1}}=\frac{\sqrt{5}}{2}$$⇒2m2+1​∣−3m∣​=25​​

$$\Rightarrow \frac{|-3m|}{\sqrt{{m}^{2}+1}}=\sqrt{5}$$⇒m2+1​∣−3m∣​=5​

Squaring both sides, we get

$$\Rightarrow 5{m}^{2}+5=9{m}^{2}$$⇒5m2+5=9m2

$$\Rightarrow 4{m}^{2}=5$$⇒4m2=5

$$\Rightarrow m=\pm \frac{\sqrt{5}}{2}.$$⇒m=±25​​.