Question

If the tangent at $P$ on $y^2=4ax$ meets the tangent at the vertex in $Q$, and $S$ is the focus of the parabola, then $\angle SQP=$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is C $\frac{\pi }{2}$

We are given $y^2=4ax$
Let any point on parabola be$P(a{t}^2,2at)$
Tangent at$P:y=\frac{x}{t}+at$
$\Rightarrow ty=x+a{t}^2\cdots \left(1\right)$
And tangent at vertex is $x=0\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$ty=a{t}^2$
$\Rightarrow y=at$
So, coordinates of $Q\equiv (0,at)\text{and}S=(a,0)$
Now, slope of $QP=\frac{2at-at}{a{t}^2-0}=\frac{1}{t}=m\left(say\right)$
Slope of $QS=\frac{0-at}{a-0}=-t=m^{\prime }$ say
Now, $m{m}^{\prime }=\frac{1}{t}\times -t=-1$
Hence $\angle SQP=\frac{\pi }{2}$