Question

Three normals are drawn from the point $\left(c,0\right)$ to the curve $y^2=x$. Show that c must be greater than $\frac{1}{2}$. One normal is always the x-axis. For what value of ${}^{\prime }{c}^{\prime }$ are the other two normals perpendicular to each other?

Answer 1

The equation of given curve is,

$y^2=x$

This is the equation of parabola. Comparing it with standard equation of parabola i.e. $y^2=4ax$, we get,

$4a=1$

$\therefore a=\frac{1}{4}$

Thus, equation of normal to the parabola is given by,

$y=mx-2am-a{m}^3$

$\therefore y=mx-2\left(\frac{1}{4}\right)m-\left(\frac{1}{4}\right)m^3$

$\therefore y=mx-\frac{m}{2}-\frac{m^3}{4}$

Normals are drawn from $(c,0)$

Thus, put $x=c$ and $y=0$ in above equation, we get,

$\therefore 0=mc-\frac{m}{2}-\frac{m^3}{4}$

$\therefore 0=m(c-\frac{1}{2}-\frac{m^2}{4})$

$\therefore m=0$ or $c-\frac{1}{2}-\frac{m^2}{4}=0$

$\therefore m_1=0$ or $c-\frac{1}{2}-\frac{m^2}{4}=0$

Thus, one normal is always x axis.

Now, $c-\frac{1}{2}-\frac{m^2}{4}=0$

$\therefore \frac{m^2}{4}=c-\frac{1}{2}$

$\therefore m^2=4(c-\frac{1}{2})$

$\therefore m_2\times m_3=4(c-\frac{1}{2})$ (1)

Now, $m^2\ge 0$

$\therefore 4(c-\frac{1}{2})\ge 0$

$\therefore c-\frac{1}{2}\ge 0$

$\therefore c\ge \frac{1}{2}$

Hence proved.

Now, if two normals are perpendicular to each other,

$m_2\times m_3=-1$

$\therefore 4(c-\frac{1}{2})=-1$

$\therefore (c-\frac{1}{2})=\frac{-1}{4}$

$\therefore c=\frac{1}{2}-\frac{1}{4}$

$\therefore c=\frac{1}{4}$