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
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
\((0,\frac{1}{4})\)
\((\frac{1}{2},\frac{3}{4})\)
\((\frac{1}{4},\frac{1}{2})\)
\((\frac{3}{4},1)\)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : \((\frac{1}{4},\frac{1}{2})\)
\(y = mx + 2a + \frac{1}{m^2}\) ( Equation of normal to x2 = 4ay in slope form)through(1,-1)
\(4m^3+6m^2+1=0\)
\(⇒ m ≃ -16\)
Slope of normal \(≃ \frac{-8}{5} = \tan \theta\)
\(⇒ \cos \theta ≃ \frac{-5}{\sqrt{89}}, \sin \theta ≃ \frac{8}{\sqrt{89}}\)
\(x_p = 1+cos \theta ≃ 1 - \frac{5}{\sqrt{89}} \in (\frac{1}{4},\frac{1}{2})\)
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Concepts Used:
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus is the theorem which states that differentiation and integration are opposite processes (or operations) of one another.
Calculus's fundamental theorem connects the notions of differentiating and integrating functions. The first portion of the theorem - the first fundamental theorem of calculus – asserts that by integrating f with a variable bound of integration, one of the antiderivatives (also known as an indefinite integral) of a function f, say F, can be derived. This implies the occurrence of antiderivatives for continuous functions.
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