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

(a) (-3/16, -3/4)

(b) (-3/16, 3/4)

(c) (3/16, -3/4)

(d) (3/16, 3/4)

This question was addressed to me in my homework.

My doubt stems from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

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The correct option is (d) (3/16, 3/4)

Easy 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 = 3×1

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

And the coordinates of the foot of the normal are (x1, y1) = (3/16, 3/4)