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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Parabola

The radius of the smallest circle which touches the parabolas...

The radius of the smallest circle which touches the parabolas y=x2+2y=x^2+2 and x=y2+2x=y^2+2 isThe radius of the smallest circle which touches the parabolas y=x^2+2 and x=y^2+2 is

Options

  1. A722\frac{7 \sqrt{2}}{2}
  2. B7216\frac{7 \sqrt{2}}{16}
  3. C724\frac{7 \sqrt{2}}{4}
  4. D728\frac{7 \sqrt{2}}{8}

Correct answer

D. 728\frac{7 \sqrt{2}}{8}

Step-by-step solution

The given parabolas are symmetric about the line y=xy=x
The radius of the smallest circle which touches the parabolas y=x^2+2 and x=y^2+2 is
Tangents at A& B\mathrm{A} \& \mathrm{~B} must be parallel to y=x\mathrm{y}=\mathrm{x} line, so slope of the tangents =1=1
\left(\frac{d y}{d x}\right)_{\min A}=1=\left(\frac{d y}{d x}\right)_{\min B}<br/>Forpoint<br/>For point\mathrm{B}, y=x\mathrm{y}=\mathrm{x}^2+2<br/><br/>\begin{gathered}
\frac{d y}{d x}=2 x=1 \\ x=\frac{1}{2} \Rightarrow y=\frac{9}{4} \\ \therefore \text{ Point } B=\left(\frac{1}{2}, \frac{9}{4}\right) \Rightarrow \text{ Point } \mathrm{A}=\left(\frac{9}{4}, \frac{1}{2}\right) \\ \mathrm{AB}=\sqrt{\left(\frac{1}{2}-\frac{9}{4}\right)^2+\left(\frac{9}{4}-\frac{1}{2}\right)^2} \\ =\sqrt{\frac{98}{16}}=\frac{7 \sqrt{2}}{4}
\end{gathered}<br/>Radius<br/>Radius==\frac{7 \sqrt{2}}{8}
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