Question

The normal chord of a parabola $y^2=4ax$ at the point whose ordinate is equal to the abscissa then angle subtended by normal chord at the focus is:

• A. $\frac{\pi }{4}$
• B. ${\tan }^{-1}\sqrt{2}$
• C. ${\tan }^{-1}2$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

Point whose ordinate equal to abscissa will be such that $x=y,$

but $(x,y)$ lies on $y^2=4ax$

$\Rightarrow$ $x^2=4ax$

$\Rightarrow$ $x=0$ or $x=4a$

Since at $(0,0)$ we cannot draw a normal chord, hence point at which normal chord is drawn is $P(4a,4a).$

Focus of the given parabola $y^2=4ax$ is $S(a,0).$

Equation of the normal chord at $(4a,4a)$ will be

$y-4\frac{a}{x}-4a=-2$

$\Rightarrow$ $2x+y=12a$ is the equation of normal.

To find the points of intersection of normal with the parabola , we need to

solve the equation $(12a-2x{)}^2=4ax$

$\Rightarrow$ $(6a-x{)}^2=ax$

$\Rightarrow$ $36a^2-12ax+x^2=ax$

$\Rightarrow$ $x^2-13ax+36a^2=0$

$\Rightarrow$ $(x-4a)(x-9a)=0$

$\Rightarrow$ $x=4a$ or $9a$

$P(4a,4a)$ and $Q(9a,-6a)$ are the end points of the normal chord.

Slope of line $PS=\frac{(4a-0)}{(4a-a)}=\frac{4}{3}$

Slope of line $QS=\frac{(-6a-0)}{(9a-a)}=\frac{-3}{4}$

We can observe that product of $\left(SlopeofPS\right)\times \left(SlopeofQS\right)=-1$

$\Rightarrow$ $\angle PSQ={90}^o$.

$\therefore$ Normal chord at the point whose ordinate is equal to abscissa subtends an angle of $\frac{\pi }{2}$ at the focus.