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 $$(x_1,y_1)$$ is given by,

$$(y-y_1)={\displaystyle \frac{-1}{{\displaystyle \frac{dy}{dx}}}}(x-x_1)$$

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 = $$(m^2,-2m)$$

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 $$(x_1,y_1)$$ is given by,

$$(y-y_1)={\displaystyle \frac{-1}{{\displaystyle \frac{dy}{dx}}}}(x-x_1)$$

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

Differentiating above equation with respect to x we get,

$$2y{\displaystyle \frac{dy}{dx}}=4$$

Dividing by 2y both sides,

$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2}{y}}$$

Here point P has $$(m^2,-2m)$$ as co – ordinate.

Then at P; $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2}{(-2m)}}$$ as y = 2m at P.

$$\Rightarrow {\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{1}{m}}$$

Substituting $$(x_1,y_1)=(m^2,-2m)$$ and $${\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{1}{m}}$$ in equation of normal of parabola we get;

$$y-(-2m)={\displaystyle \frac{-1}{{\displaystyle \frac{-1}{m}}}}(x-m^2)$$

Cancelling common negative,

$\Rightarrow y+2m=m(x-m^2)$

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

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 = $$\tan {\displaystyle \frac{\pi }{4}}$$.

And the value of $$\tan {\displaystyle \frac{\pi }{4}}$$ = 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 $$P=(m^2,-2m)$$,

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

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 $$(m^2,+2m)$$ or $$(m^2,-2m)$$. 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.