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 and is
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
The given parabolas are symmetric about the line

Tangents at must be parallel to line, so slope of the tangents
\left(\frac{d y}{d x}\right)_{\min A}=1=\left(\frac{d y}{d x}\right)_{\min B}\mathrm{B}, ^2+2\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}==\frac{7 \sqrt{2}}{8}
Tangents at must be parallel to line, so slope of the tangents
\left(\frac{d y}{d x}\right)_{\min A}=1=\left(\frac{d y}{d x}\right)_{\min B}\mathrm{B}, ^2+2\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}==\frac{7 \sqrt{2}}{8}