Question

If the three normals drawn to the parabola, $y^2=2x$ pass through the point $\left(a,0\right),a\ne 0$, then ‘$a$’ must be greater than

• A. $1$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is A

$1$

Explanation for the correct option.

Step 1. Finding the equation of the normal:

The equation of the normal drawn to the parabola of the form $y^2=4ax$ is given as $y=mx-2am-a{m}^3$.

For the parabola $y^2=2x$, the value of constant $a$ is given by comparing the equation with the standard form:

$\begin{array}{rcl}4a& =& 2\\ & \Rightarrow & a=\frac{2}{4}\\ & \Rightarrow & a=\frac{1}{2}\end{array}$

So the equation of normal drawn to parabola $y^2=2x$ is given by substituting $a=\frac{1}{2}$ in the equation $y=mx-2am-a{m}^3$:

$\begin{array}{rcl}& \Rightarrow & y=mx-2\times \frac{1}{2}m-\frac{1}{2}{m}^3\\ & \Rightarrow & y=mx-m-\frac{1}{2}{m}^3\end{array}$

Step 2. Substitute the point in the equation and find the value of $a$.

As the point $(a,0)$ passes through the normal, so substitute the point in the equation of the normal.

$\begin{array}{rcl}0& =& ma-m-\frac{1}{2}{m}^3\\ & \Rightarrow & 0=2ma-2m-m^3\\ & \Rightarrow & 0=m(2a-2-m^2)\end{array}$

So either $m=0$ or

$\begin{array}{rcl}& \Rightarrow & 2a-2-m^2=0\\ & \Rightarrow & 2a-2=m^2\end{array}$

Now, for the parabola to have three normals $m^2>0$ so:

$\begin{array}{rcl}2a-2& >& 0\\ & \Rightarrow & 2a-2+2>0+2\\ & \Rightarrow & 2a>2\\ & \Rightarrow & a>\frac{2}{2}\\ & \Rightarrow & a>1\end{array}$

Hence, the correct option is A.