JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q81

← JEE Main 2014 · 19 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 81 of 90

A chord is drawn through the focus of the parabola $y^2=6x$ such that its distance from the vertex of this parabola is $\dfrac{\sqrt5}2$ , then its slope can be :

\dfrac{\sqrt5}2

\dfrac{\sqrt3}2

\dfrac2{\sqrt5}

\dfrac2{\sqrt3}

Answer and worked solution

Official answer: A

The focus is $\left(\frac32,0\right)$ . The chord $y=m\left(x-\frac32\right)$ is at distance $\dfrac{\frac32|m|}{\sqrt{1+m^2}}=\dfrac{\sqrt5}2$ from the vertex.

So $9m^2=5(1+m^2)$ , $m^2=\frac54$ and $m=\dfrac{\sqrt5}2$ .

Topics:Parabola Distance of a point from a line

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 81