Question

What will be the equation of the normal to the parabola y^2 = 3x which is perpendicular to the line y = 2x + 4?

(a) 16x + 32y = 27

(b) 16x – 32y = 27

(c) 16x + 32y = -27

(d) -16x + 32y = 27

I had been asked this question by my college professor while I was bunking the class.

The query is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

Answer 1

The correct answer is (a) 16x + 32y = 27

Best explanation: Given, y^2 = 3x ……….(1)  and y = 2x + 4  ……….(2)

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

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x1, y1) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]P = -2y1/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y1/3*2 = -1

Since the slope of the line (2) is 2

Or y1 = 3/4

Since the point P(x1, y1) lies on (1) hence,

y1^2 = 3x1

As, y1 = 3/4, so, x1 = 3/16

Therefore, the required equation of the normal is

y – y1 = -(2y1)/3*(x – x1)

Putting the value of x1 and y1 in the above equation we get,

16x + 32y = 27.