Question

Two tangents are drawn to the parabola $y^2=8x$ which meets the tangent at vertex at $P$ and $Q$ respectively. If $PQ=4\text{units}$, then the locus of the point of intersection of the two tangents is

• A. $y^2=8(x+2)$
• B. $y^2=8(x-2)$
• C. $x^2=8(y+2)$
• D. $x^2=8(y-2)$

Answer 1

The correct option is A $y^2=8(x+2)$

$S:y^2=8x\Rightarrow a=2$
Let the two points on parabola be $R\left(t_1\right)$ and $S\left(t_2\right)$


Equation of the tangent at the vertex is
$x=0$
Equation of tangent at $R\left(t_1\right)$
$t_{1}y=x+a{t}_1^2\Rightarrow P\equiv (0,a{t}_1)$
Similarly, we get
$Q\equiv (0,a{t}_2)$
Now,
$\Rightarrow PQ=|a{t}_1-a{t}_2|=4\Rightarrow |t_1-t_2|=2(\because a=2)\Rightarrow (t_1-t_2{)}^2=4\cdots (i)$

Point of intersection of tangent at $R$ and $S$ is
$T(h,k)=(a{t}_1{t}_2,a(t_1+t_2\left)\right)\Rightarrow t_1{t}_2=\frac{h}{a},t_1+t_2=\frac{k}{a}\Rightarrow (t_1-t_2{)}^2=(t_1+t_2{)}^2-4t_1{t}_2\Rightarrow 4=\frac{k^2}{a^2}-\frac{4h}{a}\Rightarrow 4=\frac{k^2}{4}-\frac{4h}{2}(\because a=2)$

Hence, the locus is
$y^2=8(x+2)$