Question

At what point on the parabola ${{y}^{2}}=4x$ the normal makes an equal angle with the axis?
(a) (4, 4)
(b) (9, 6)
(c) (4, -4)
(d) (1, $\pm$2)

Answer 1

Hint: To solve this question we will first of all determine the equation of normal of the given parabola. The equation of normal of parabola of type, ${{y}^{2}}=4x$ an point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by,

$\left( y-{{y}_{1}} \right)=\dfrac{-1}{\dfrac{dy}{dx}}\left( x-{{x}_{1}} \right)$

After obtaining the equation of normal and assuming co – ordinates of point P we will try to determine the value of slope m of normal. Thus, it will help us to get the value of co – ordinates of point P.

Complete step-by-step solution:

Let us assume the point on the parabola is P.

We will first assume the coordinates of P.

Let x – coordinate of P be ${{m}^{2}}$, then as P lies on the parabola ${{y}^{2}}=4x$. So, it must satisfy the equation ${{y}^{2}}=4x$.

Substituting $x={{m}^{2}}$ as ${{y}^{2}}=4x$ to get y – coordinate of P we get,

${{y}^{2}}=4\left( {{m}^{2}} \right)$

Taking square roots on both sides we get,

$y=\pm 2m$

So we can consider the y – coordinate of P as +2m or -2m.

Let it be -2m.

When P = $\left( {{m}^{2}},-2m \right)$

So, we have a figure as,

Now we have to consider normal at P.

The equation of normal of parabola of type, ${{y}^{2}}=4x$ an point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by,

$\left( y-{{y}_{1}} \right)=\dfrac{-1}{\dfrac{dy}{dx}}\left( x-{{x}_{1}} \right)$

Given ${{y}^{2}}=4x$, we will calculate $\dfrac{dy}{dx}$ now,

Differentiating above equation with respect to x we get,

$2y\dfrac{dy}{dx}=4$

Dividing by 2y both sides,

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{y}$

Here point P has $\left( {{m}^{2}},-2m \right)$ as co – ordinate.

Then at P; $\dfrac{dy}{dx}=\dfrac{2}{\left( -2m \right)}$ as y = 2m at P.

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{m}$

Substituting $\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{m}^{2}},-2m \right)$ and $\dfrac{dy}{dx}=-\dfrac{1}{m}$ in equation of normal of parabola we get;

$y-\left( -2m \right)=\dfrac{-1}{\dfrac{-1}{m}}\left( x-{{m}^{2}} \right)$

Cancelling common negative,

$\Rightarrow y+2m=m\left( x-{{m}^{2}} \right)$

$\Rightarrow y+2m=mx-{{m}^{3}}$

Subtracting 2m both sides,

$\Rightarrow y=mx-{{m}^{3}}-2m$ - (1)

This is the equation of normal.

Given that normal makes an equal angle with the axis.

The slope = m = $\tan \dfrac{\pi }{4}$.

And the value of $\tan \dfrac{\pi }{4}$ = 1.

$\Rightarrow m=1$

Substituting m = 1 in equation (1) we get,

$\Rightarrow y=x-1-2$

$\Rightarrow y=x-3$

Now finally we have to calculate $P=\left( {{m}^{2}},-2m \right)$,

$\Rightarrow P=\left( +1,-2 \right)$

So the point is (1, -2) and it is (1, +2) when P is taken as (-y, +2m).

So option (a) is correct.

Note: Student may get confused while assuming co – ordinates of point P at $\left( {{m}^{2}},+2m \right)$ or $\left( {{m}^{2}},-2m \right)$. Both are correct, you can proceed for selecting any one of above as co – ordinate of P and then proceed for solution. Finally at the end you can use the other left one to get the full solution.