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 = 27I 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

What will be the equation of the normal to the parabola y^2 = 3x which

1.	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 = 27I 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» 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, -2y1/32 = -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.

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