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

Category: QuestionsWhat will be the equation of the normal to the parabola y^2 = 3x which is perpendicular to the line y = 2x + 4?
Editor">Editor Staff asked 11 months ago

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
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1 Answers
Editor">Editor Staff answered 11 months ago

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 = 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.