Question

The locus of point $P$ when three normals drawn from it to parabola $y^2=4ax$ are such that two of them make complementary angles is

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

Answer 1

The correct option is A $y^2=a(x-a)$

Eqaution of normal to $y^2=4ax$ passing through $P$ whose locus is $(h,k)$ is
$k=mh-2am-a{m}^3$
$\Rightarrow a{m}^3+(2a-h)m+k=0\cdots \left(1\right)$
It has $3$ roots.
$m_1{m}_2{m}_3=\frac{-k}{a}$
As $2$ tangents make complementary angles,
so if $m_1=\tan \theta ,$ then $m_2=\cot \theta$
$\therefore m_3=\frac{-k}{a}$
Putting this in equation $\left(1\right),$
$a{\left(\frac{-k}{a}\right)}^3+(2a-h)\left(\frac{-k}{a}\right)+k=0$
By solving, we get
$k^2=ah-a^2$
$\therefore y^2=a(x-a)$