Question

If $P$ is a point on the parabola $y^2=4x$ in which the abscissa is equal to ordinate then the equation of the normal at $P$ is

• A. $2x+y+12a=0$
• B. $2x+y-12a=0$
• C. $2x+y-18a=0$
• D. $x+2y-12a=0$

Answer 1

The correct option is B $2x+y-12a=0$

Since $p$ line on parabola and its coordinates satisfies

$y=x$

putting $y=x$ in the equation

$\begin{array}{c}{y}^2=4x\\ x^2=4x\\ x^2-4x=0\\ x\left(x-4\right)=0\\ x=0,4\\ so\\ y=0,4\end{array}$

therefore point $p$ is $\left(0,0\right)$ or $\left(4,4\right)$

Now

$\begin{array}{c}{y}^2=4x\\ 2y\frac{dy}{dx}=4\\ \frac{dy}{dx}=\frac{2}{y}\\ \left(\frac{dy}{dx}\right)\left(0,0\right)=\infty \\ \left(\frac{dy}{dx}\right)\left(4,4\right)=\frac{2}{4}=\frac{1}{2}\end{array}$

equation of normal at $\left(0,0,\right)$ is

$\begin{array}{c}y-0=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(0,0,\right)}}\left(x-0\right)\\ y=\frac{-1}{\infty }x\\ y=0\end{array}$

equation of normal at $\left(4,4\right)$ is

$\begin{array}{c}y-4=\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(4,4\right)}}\left(x-4\right)\\ y-4=\frac{-1}{\frac{1}{2}}\left(x-4\right)\\ y-4=-2\left(x-4\right)\\ y-4=-2x+8\\ 2x+y-12=0\end{array}$

thus the required equation of normal is $2x+y-12a=0$

$B$ is the correct answer