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

What will be the co-ordinates of the foot of the normal to the parabol

1.	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
Answer» 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, -2y1/32 = -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 And the coordinates of the foot of the normal are (x1, y1) = (3/16, 3/4)

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