Question

If point (h, k) lies on the axis of the parabola $y^2=4ax$, then find the condition for point (h, k) sothat exactly three normal can be drawn.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

LEt the equation of the normal be y=mx+c

Equation of the tangent on the parabola $y^2=4axatP(x_1,y_1)$ is

$y{y}_1=2a(x+x_1)$

$y=\frac{2a}{y_1}x+\frac{2a}{y_1}{x}_1$

Slope of tangent $=\frac{2a}{y_1}$

Slope of the normal =$\frac{-1}{m_1}=\frac{-1}{\frac{2a}{y_1}}=\frac{-y_1}{2a}$

Slope of the normal m=$\frac{-y_1}{2a}\Rightarrow y_1=-2am$

$y^2=4ax$

$P(x_1,y_1)$ should satisfy the parabola

$y_1^2=4a{x}_1$

Substituting $y_1=-2am$

We get, $(-2am{)}^2=4a{x}_1$

$4a^2{m}^2=4a{x}_1$

$x_1=a{m}^2$

Coordinates of point $P(a{m}^2,-2am)$ where m is the slope of the normal

Let equation of normal $y=mx+c$

Point P should satisfy the equation of the normal

$-2am=m\left(a{m}^2\right)+c$

$c=-a{m}^3-2am$

Equation of the normal being

$y=mx-2am-a{m}^3$ .........(1)

This normal passes through the point (h,k) where point (h,k) lies on

the axis of the parabola

$\Rightarrow$ y-coordinate=0,k=0

Substituting point (h,k) on the equation of the normal

$0=hm-2am-a{m}^3$

$m(h-2a-am2)=0$

$m=0orh-2a=a{m}^2$

$m^2=\frac{h-2a}{a}$

We need to find the condition for which exactly three normal can be drawn.

So, we need three different values of m

Already we got,

$m=0and{m}^2=\frac{h-2a}{a}$

$m=0$ means it's x-axis

To get three different normal

$m^2$ should be greater than zero

Already m=0 gives one normal

$\frac{h-2a}{a}>0$ {a is position for standard parabola}

$h-2a>0$

$h>2a$

we have to take point on the parabola such that x coordinates should be more than the semi lactus rectum of the parabola.

So,If $h>2a$ we can draw exactly three normal to the parabola.