Question

If the tangent and normals at the extremities of a focal chord of a parabola intersect at $(x_1,y_1)$ and $(x_2,y_2)$ respectively, then

• A. $x_1=x_2$
• B. $x_1=y_2$
• C. $y_1=y_2$
• D. $y_1=x_2$

Answer 1

The correct option is C $y_1=y_2$

Let $P(a{t_1}^2,2a{t}_1)$ and $Q(a{t_2}^2,2a{t}_2)$ be a focal chord of the parabola $y^2=4ax$.

For a focal chord, $t_1{t}_2=-1$
The tangents at $P$ and $Q$ intersect at $(a{t}_1{t}_2,a(t_1+t_2))$
$\therefore x_1=a{t}_1{t}_2$ and $y_1=a(t_1+t_2)$
$\Rightarrow x_1=-a$ and $y_1=a(t_1+t_2)$ $[\because PQ$ is a focal chord, $\therefore t_1{t}_2=-1]$
The normals at $P$ and $Q$ intersect at $(2a+a({t_1}^2+{t_2}^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2))$
$\therefore x_2=2a+a({t_1}^2+{t_2}^2+t_1{t}_2)$ and $y_2=-a{t}_1{t}_2(t_1+t_2)$
$\Rightarrow x_2=2a+a({t_1}^2+{t_2}^2-1)=a+a({t_1}^2+{t_2}^2)$ and $y_2=a(t_1+t_2)$
Hence $y_1=y_2$