What will be the equation of the normal to the parabola y^2 = 5x that makes an angle 45° with the x axis?

(a) 4(x – y) = 15

(b) 4(x + y) = 15

(c) 2(x – y) = 15

(d) 2(x + y) = 15

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This interesting question is from Calculus Application in section Application of Calculus of Mathematics – Class 12

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Right answer is (a) 4(x – y) = 15

The best I can explain: 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