Question

The point of intersection of tangents drawn at the end points of latus rectum of the parabola $6x^2=4y$ is

• A. $(0,\frac{-1}{6})$
• B. $(-2,0)$
• C. $(0,-3)$
• D. None of these

Answer 1

The correct option is A $(0,\frac{-1}{6})$

Given parabola is $x^2=4ay$ (where $a=\frac{1}{6}$)
So coordinates of the end point of its latus rectum are $P_1(2a,a)$ and $P_2(-2a,a)$
Now slope of tangent passes through $P_1$ is $=\left(\frac{1}{2a}\right)\left(2a\right)=1$
and slope of tangent passes through $P_2$ is $=\left(\frac{1}{2a}\right)(-2a)=-1$
Hence equation of tangent through $P_1$ is $(y-a)=1(x-2a)..\left(1\right)$
and equation of tangent through $P_2$ is $(y-a)=-1(x+2a)..\left(2\right)$
solving (1) and (2) we get point of intersection $(0,-a)=(0,-\frac{1}{6})$