What will be the co-ordinates of the foot of the normal to the parabola y^2 = 5x that makes an angle 45° with the x axis?(a) (-5/4, 5/2)(b) (5/4, 5/2)(c) (5/4, -5/2)(d) (-5/4, -5/2)I have been asked this question in an international level competition.My query is from Calculus Application in section 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 = 5x that makes an angle 45° with the x axis?(a) (-5/4, 5/2)(b) (5/4, 5/2)(c) (5/4, -5/2)(d) (-5/4, -5/2)I have been asked this question in an international level competition.My query is from Calculus Application in section Application of Calculus of Mathematics – Class 12
Answer» The correct answer is (B) (5/4, 5/2) Easiest explanation: The equation of the given PARABOLA is, y^2 = 5X ……….(1) Differentiating both SIDES of (1) with respect to y, we get, 2y = 5(dx/dy) Or dx/dy = 2y/5 Take any point P((5/4)t^2, (5/2)t). Then, the normal to the curve (1) at P is, -[dx/dy]P = -(2*5t/2)/5 = -t By the question, slope of the normal to the curve (1) at P is tan45°. Thus, -t = 1 Or t = -1 So, the REQUIRED equation of normal is, y – 5t/2 = -t(x – 5t^2/4) Simplifying further we get, 4(x – y) = 15 The co-ordinates of the foot of the normal are, P((5/4)t^2, (5/2)t). As t = 1, so putting the value of t = 1, we get, P = (5/4, 5/2).

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