Question

If a tangent to the parabola $y^2=8x$ meets the $x$-axis at $T$ and intersect the tangent at vertex $A$ at $P$, and the rectangle $TAPQ$ is completed, then the locus of the point $Q$ is

• A. $y^2+2x=0$
• B. $x^2+2y=0$
• C. $x^2-2y=0$
• D. $y^2-2x=0$

Answer 1

The correct option is A $y^2+2x=0$


$y^2=8x$
$\Rightarrow a=2$
Vertex $A$ is $(0,0)$.
The tangent at any point $(a{t}^2,2at)$ of the parabola $y^2=4ax$ is given by $ty=x+a{t}^2$.
Therefore, the tangent at any point of $y^2=8x$ is $ty=x+2t^2$.
Since, the tangent at vertex $A$ is $y$-axis,
so, $T$ and $P$ are $(-2t^2,0)$ and $(0,2t)$, respectively.

Let point $Q$ be $(h,k)$.
Then $h=AT=-2t^2\cdots \left(1\right)$
and $k=AP=2t\cdots \left(2\right)$
Eliminating $t$, using $\left(1\right)$ and $\left(2\right)$, we get
$2h=-(k{)}^2$
$\Rightarrow k^2+2h=0$
Hence, the locus of $Q$ is $y^2+2x=0$