Question

The point of intersection of normals to the parabola $y^2=4x$ at the points whose ordinates are $4$ and $6$ is

• A. $(30,-21)$
• B. $(21,-30)$
• C. $(17,-19)$
• D. $(19,-18)$

Answer 1

The correct option is A $(30,-21)$

If $y=4$

$y^2=4x$

$x=4$

If $y=6$

then,

$x=9$

So, points are $(4,4),(9,6)$

Now,

$y^2=4x$

$2y\frac{dy}{dx}=4$

$\frac{dy}{dx}=\frac{2}{y}$

Equation of the normal at point $(4,4)$ is

$y-4=-\frac{1}{(\frac{dy}{dx}{)}_{(4,4)}}(x-4)$

$y-4=-\frac{1}{\left(\frac{2}{4}\right)}(x-4)$

$2x+y=12$ ----- $\left(1\right)$

Equation of the normal at $(9,6)$ is

$y-6=-\frac{1}{(\frac{dy}{dx}{)}_{(9,6)}}(x-9)$

$y-6=-\frac{1}{\left(\frac{2}{6}\right)}(x-9)$

$3x+y=33$ ---- $\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$

$-x=-21$

$x=21$

putting $x=21$ in $\left(1\right)$

$2x+y=12$

$42+y=12$

$y=-30$

So, the point of intersection of the two normal is $(21,-30)$

Hence, the option $B$ is the correct answer.