Question

The locus of the centre of the circle described on any focal chord of the parabola $y^2=4ax$ as the diameter is

• A. $y^2=2a\left(x+a\right)$
• B. $y^2=a\left(x+a\right)$
• C. $y^2=2a\left(x-a\right)$
• D. $y^2=4a\left(x-a\right)$

Answer 1

Answer: (C)

Let, $P\left(a{t}_1^2,2a{t}_1\right)$ and $Q\left(a{t}_2^2,2a{t}_2\right)$ be the extremities of a focal chord $PQ$ of the parabola $y^2=4ax.$ Then, $t_1{t}_2=-1.$
Let, $\left(h,k\right)$ be the coordinates of the centre of the circle described on $PQ$ as diameter. Then,
$h=\frac{a}{2}\left(t_1^2+t_2^2\right)$ and $k=a\left(t_1+t_2\right)$
$\Rightarrow \frac{2h}{a}=t_1^2+t_2^2$ and ${\left(\frac{k}{a}\right)}^2={\left(t_1+t_2\right)}^2$
$\Rightarrow \frac{2h}{a}=t_1^2+t_2^2$ and $\frac{k^2}{a^2}=t_1^2+t_2^2+2t_1{t}_2$
$\Rightarrow \frac{k^2}{a^2}=\frac{2h}{a}-2$ $\left[\because t_1{t}_2=-1\right]$
$\Rightarrow k^2=2a\left(h-a\right)$
Hence, the locus of $\left(h,k\right)$ is
$y^2=2a\left(x-a\right).$