Question

At what point on the parabola y^2=4x the normal makes equal angles with the axes

Answer 1

The given equation of parabola is ,$y^{2} = 4 x$

Let that point be (p,q).

Equation of tangent at (p,q) is given by, 2 y y'=4 where y' is derivative of y.

→y' = 2/ y

→y'(p,q)= 2/q

Slope of normal = -1/y'=-q/2

As given normal makes equal angle with axes.

Equation of axes is , x y=0

Differentiating w.r.t x, we get

y + x y' =0

y' = -y/x

y'(p,q)=-q/p

So, slope of normal and slope of axes should be same.

→ -q/p=-q/2

→ p = 2 .

As point (p,q) lies on parabola, y² =4 x, so q²= 4 p .........(1)

putting the value of p in (1)

→q²=8

→ q =$\pm2{\sqrt{2}}$

∵ the point on the parabola y²=4 x is (2,√2) and (2,-√2) at which the normal makes equal angles with the axes.