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

Category: QuestionsWhat will be the equation of the normal to the parabola y^2 = 5x that makes an angle 45° with the x axis?
Editor">Editor Staff asked 11 months ago

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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1 Answers
Editor">Editor Staff answered 11 months ago

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