QJEEM22Ma67e5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma67e5 Solution: Let P and Q be any points on the curves ( x − 1 ) 2 + ( y + 1 ) …

Question

Verified Solution • 2025–26 Session moderate

QJEEM22Ma67e5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma67e5 Solution: Let P and Q be any points on the curves ( x − 1 ) 2 + ( y + 1 ) …

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

Multiple Choice Options

A (0,14)
B (12,34)
C (14,12)
D (34,1)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring

Accepted Answer

$y = m x + 2 a + \frac{1}{m^{2}}$ (Equation of normal to $x^{2} = 4 a y$ in slope form) through $(1, - 1)$ .

$4 m^{3} + 6 m^{2} + 1 = 0$

$\Rightarrow m ≃ - 1.6$

Slope of normal $≃ \frac{- 8}{5} = \tan θ$

$\Rightarrow \cos θ ≃ \frac{- 5}{\sqrt{89}}, \sin θ ≃ \frac{8}{\sqrt{89}}$

$x_{p} = 1 + \cos θ ≃ 1 - \frac{5}{\sqrt{89}} \in (\frac{1}{4}, \frac{1}{2})$

QJEEM22Ma67e5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma67e5 Solution: Let P and Q be any points on the curves ( x − 1 ) 2 + ( y + 1 ) …

Multiple Choice Options

Step-by-Step Verified Solution

Related Learning Resources & Exam Preparation