Question

A chord is drawn through the focus of the parabola $y^2=6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt{5}}{2}$, then its slope can be :-

• A. $\frac{\sqrt{5}}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (A)

Equation of parabola, $y^2=6x$
$\Rightarrow v^2=4\times \frac{3}{2}x$
$\therefore$ Focus $=\left(\frac{3}{2},0\right)$
Let equation of chord passing through focus be
$ax+by+c=0\dots \left(1\right)$
Since chord is passing through $\left(\frac{3}{2},0\right)$
$\therefore$ Put $x=\frac{3}{2},y=0$ in eqn $\left(1\right)$, we get
$\frac{3}{2}a+c=0$
$\Rightarrow c=-\frac{3}{2}a...\left(2\right)$
distance of chord from origin is $\frac{\sqrt{5}}{2}$
$\frac{\sqrt{5}}{2}=\left|\frac{a\left(0\right)+b\left(0\right)+c}{\sqrt{a^2+b^2}}\right|=\frac{c}{\sqrt{a^2+b^2}}$
Squaring both sides
$\frac{5}{4}=\frac{c^2}{a^2+b^2}$
$\Rightarrow a^2+b^2=\frac{4}{5}{c}^2$
Putting value of c from $\left(2\right)$, we get
$a^2+b^2=\frac{4}{5}\times \frac{9}{4}{a}^2$
$b^2=\frac{9}{4}{a}^2-a^2=\frac{4}{5}{a}^2$
$\frac{a^2}{b^2}=\frac{5}{4},\frac{a}{b}=\pm \frac{\sqrt{5}}{2}$
Slope of chord, $\frac{dy}{dx}$
$=-\frac{a}{b}=-\left(\frac{\pm \sqrt{5}}{2}\right)=\mp$
$\frac{\sqrt{5}}{2}$