Question

What will be the co-ordinates of the foot of the normal to the parabola y^2 = 5x that makes an angle 45° with the x axis?

(a) (-5/4, 5/2)

(b) (5/4, 5/2)

(c) (5/4, -5/2)

(d) (-5/4, -5/2)

I have been asked this question in an international level competition.

My query is from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer 1

The correct answer is (b) (5/4, 5/2)

Easiest explanation: The equation of the given parabola is, y^2 = 5x ……….(1)

Differentiating both sides of (1) with respect to y, we get,

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t^2, (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]P = -(2*5t/2)/5 = -t

By the question, slope of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the required equation of normal is,

y – 5t/2 = -t(x – 5t^2/4)

Simplifying further we get,

4(x – y) = 15

The co-ordinates of the foot of the normal are, P((5/4)t^2, (5/2)t).

As t = 1, so putting the value of t = 1, we get,

P = (5/4, 5/2).