Question

If the locus of the circumcentre of of variable triangle having sides $y-$axis, $y=2$ and $lx+my=1$, where $(l,m)$ lies on the parabola $y^2=4ax$ is a curve $C$, then the curve $C$ is symmetric about the line

• A. $y=-\frac{3}{2}$
• B. $y=\frac{3}{2}$
• C. $x=-\frac{3}{2}$
• D. $x=\frac{3}{2}$

Answer 1

The correct option is B $y=\frac{3}{2}$

In the above diagram,
$C\equiv (0,\frac{1}{m})$
$B=(\frac{1-2m}{l},2)$
since from the figure it is clear required triangle will be a rightangle triangle.
Let $(h,k)$ be the circumcentre of $△ABC$.
$\therefore h=\frac{1-2m}{2l}$,
$k=\frac{1+2m}{2m}\Rightarrow m=\frac{1}{2(k-1)}$
$\therefore l=\frac{k-2}{2h(k-1)}$,

$\because (l,m)$ lies on $y^2=4ax$
$\therefore m^2=4al$
$\Rightarrow {\left(\frac{1}{2(k-1)}\right)}^2=4a\left\{\frac{k-2}{2h(k-1)}\right\}$
$h=8a(k^2-3k+2)$
$\therefore$ Locus of $(h,k)$ is $x=8a(y^2-3y+2)$
$\Rightarrow {(y-\frac{3}{2})}^2=\frac{1}{8a}(x+2a)$
$\therefore$ Vertex is $(-2a,\frac{3}{2})$.
$\because$ Length of smallest focal chord
$=$ Length of latusrectum $=\frac{1}{8a}$
From the equation of curve $C$ it is clear that it is symmetric about line $y=\frac{3}{2}$