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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Application of Derivatives

The shortest distance between the curves y^2=8 x and x^2+y^2+12...

The shortest distance between the curves y2=8xy^2=8 \mathrm{x} and x2+y2+12y+35=0x^2+y^2+12 y+35=0 is :

Options

  1. A23−12 \sqrt{3}-1
  2. B2\sqrt{2}
  3. C32−13 \sqrt{2}-1
  4. D22−12 \sqrt{2}-1

Correct answer

D. 22−12 \sqrt{2}-1

Step-by-step solution

Equation of normal to parabolay2=8x is y=mx−4m−2m3y^2=8 x \text{ is } y=m x-4 m-2 m^3passes through (0,−6)(0,-6) we get−6=−4m−2m3⇒m3+2m−3=0⇒(m−1)(m2+m+3)=0⇒m=−1P=(am2,−2am)=(2,−4)∴ Shortest distance =PC−r=(22−1)\begin{aligned}& -6=-4 \mathrm{ m}-2 \mathrm{ m}^3 \\ & \Rightarrow \mathrm{ m}^3+2 \mathrm{ m}-3=0 \\ & \Rightarrow(\mathrm{ m}-1)\left(\mathrm{m}^2+\mathrm{m}+3\right)=0 \Rightarrow \mathrm{ m}=-1 \\ & \mathrm{P}=\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(2,-4) \\ & \therefore \text{ Shortest distance }=\mathrm{PC}-\mathrm{r} \\ & =(2 \sqrt{2}-1)\end{aligned}
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