Question

If two normals to a parabola $y^2=4ax$ intersect at right angles, then the chord joining their feet passes through a fixed point whose coordinates are

• A. $(-2a,0)$
• B. $(a,0)$
• C. $(2a,0)$
• D. $(-a,0)$

Answer 1

The correct option is B $(a,0)$

Given parabola is $y^2=4ax$
Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ be two points on the parabola.
Slope of normal at $P$ is $m_1=-t_1$
Slope of normal at $Q$ is $m_2=-t_2$
$\because$ normals are perpendicular to each other,
$\therefore m_1{m}_2=-1\Rightarrow t_1{t}_2=-1$
which is the required condition for focal chord.
So, chord passes through the fixed point $(a,0)$