Looks pretty normal to me

The given point (h,k) is the co-normal point w.r.t parabola $y^{2} = 4 \times \ \frac{1}{100} \times x$ .

We know that the equation of any normal to a general parabola is of the form $y = mx -2am -am^{3}$ .

Also it passes through (h,k) . So $k = mh -2am -am^{3}$ ; which comes out to be $am^{3} + m(2a-h) +k =0$ (Equation 1)

Since this is a cubic in m, therefore it has roots m 1 , m 2 and m_3 .

By Vieta's we get :

• m 1 +m 2 +m_3 = 0

• $m_1 . m_2 +m_2 . m_3 +m_3 . m_1 = \frac{2a-h}{a}$

• $m_1 . m_2 . m_3 = \frac{-k}{a}$

We are given that the three normals exist , so then all the three roots of the above Equation 1 are real and hence we can say that -

$m_1^{2} +m_2^{2} +m_3^{2} \geq 0$

$(m_1 +m_2 +m_3)^{2} - 2 . (m_1 . m_2 +m_2 . m_3 +m_3 . m_1) \geq 0$

We know that by applying Vieta's Formulae in Equation 1 ;

• $m_1 +m_2 +m_3 =0$ and

• $m_1 . m_2 +m_2 . m_3 +m_3 . m_1 = \frac{2a-h}{a}$

Therefore we get $h-2a \geq 0$ Implies that $h\geq 2a$

So, now we supply the value of $a =\frac{1}{100}$ and Voila !!!

Answer = $h \geq 0.05$ .

The equation of normal at general point $(a{t}^{2},2at)$ of parabola ${y}^{2}=4ax$ is $y+xt=2at+a{t}^{3}$ .

Since $(h,k$ lies on the normal so it must satisfy the equation of normal.

So $k+ht=2at+a{t}^{3}$ .

Now there are three different points on parabola which satisfy above equation. So there must be three different roots of the equation. And after differentiating it w.r.t the number of roots will decrease to two.

So, the equation $3a{t}^{2}+2a-h=0$ have two different roots.

Therefore $h\ge 2a$ .

Now coming to this question, here $a=\frac{1}{100}$ .

then $h\ge \frac{2}{100}$ or $h\ge 0.02$

This is a very common question that we encounter in jee preparation