Solveeit Logo

Question

Mathematics Question on Fundamental Theorem of Calculus

Let P and Q be any points on the curves (x – 1)2 + (y + 1)2 = 1 and y = x2, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval

A

(0,14)(0,\frac{1}{4})

B

(12,34)(\frac{1}{2},\frac{3}{4})

C

(14,12)(\frac{1}{4},\frac{1}{2})

D

(34,1)(\frac{3}{4},1)

Answer

(14,12)(\frac{1}{4},\frac{1}{2})

Explanation

Solution

The correct answer is (C) : (14,12)(\frac{1}{4},\frac{1}{2})
y=mx+2a+1m2y = mx + 2a + \frac{1}{m^2} ( Equation of normal to x2 = 4ay in slope form)through(1,-1)
4m3+6m2+1=04m^3+6m^2+1=0
m16⇒ m ≃ -16
Slope of normal 85=tanθ≃ \frac{-8}{5} = \tan \theta
cosθ589,sinθ889⇒ \cos \theta ≃ \frac{-5}{\sqrt{89}}, \sin \theta ≃ \frac{8}{\sqrt{89}}
xp=1+cosθ1589(14,12)x_p = 1+cos \theta ≃ 1 - \frac{5}{\sqrt{89}} \in (\frac{1}{4},\frac{1}{2})