Question

The length of normal chord to the parabola $y^2=4x$ which subtends a right angle at the vertex is

• A. $6\sqrt{3}$
• B. $6\sqrt{2}$
• C. $7\sqrt{2}$
• D. $7\sqrt{3}$

Answer 1

The correct option is A $6\sqrt{3}$

A chord is a line segment that passes through any two points on the parabola. A normal chord is a chord that is perpendicular to a tangent of the parabola at the point of intersection of the chord with the parabola.

Let $y=mx+c$ is the tangent of parabola.

As given parabola is $y^2=4x$

So any point on the parabola is $(t^2,2t)$

As $y=mx+\frac{a}{m}$ , is the tangent to the parabola $y^2=4ax$

then the tangent equation to this given parabola is $y=mx+\frac{1}{m}$

Let the tangent passes through point $P(t_1^2,2t_1)$

On substituting P in parabola equation we get $m=\frac{1}{t_1}$

So the slope of the normal is $-t_1$

Let the chord joins $P(t_1^2,2t_1)$ and $Q(t_2^2,2t_2)$

On solving we will get slope of line PQ as $\frac{2}{t_1+t_2}$

So , $\frac{2}{t_1+t_2}=\frac{1}{t_1}$

$\Rightarrow t_1^2+t_1{t}_2=-2$

As from properties of a normal chord which subtends a right angle at the vertex, $t_1{t}_2=-4$

On solving above two equations we get $t_1=\sqrt{2},t_2=-2\sqrt{2}$

Hence the points are $P(2,2\sqrt{2})$ and $Q(8,-4\sqrt{2})$

By applying distance formula we get the distance between P and Q as

$\Rightarrow PQ=\sqrt{(8-2{)}^2+(-4\sqrt{2}-2\sqrt{2}{)}^2}$

$\Rightarrow PQ=\sqrt{108}=6\sqrt{3}units$