JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The radius of the smallest circle which touches the parabolas \(y=x^2+2\) and \(x=y^2+2\) is
- A \(\frac{7 \sqrt{2}}{2}\)
- B \(\frac{7 \sqrt{2}}{16}\)
- C \(\frac{7 \sqrt{2}}{4}\)
- D \(\frac{7 \sqrt{2}}{8}\)
Answer & Solution
Correct Answer
(D) \(\frac{7 \sqrt{2}}{8}\)
Step-by-step Solution
Detailed explanation
The given parabolas are symmetric about the line \(y=x\) Tangents at \(\mathrm{A} \& \mathrm{~B}\) must be parallel to \(\mathrm{y}=\mathrm{x}\) line, so slope of the tangents \(=1\) \(\left(\frac{d y}{d x}\right)_{\min A}=1=\left(\frac{d y}{d x}\right)_{\min B}\) For point…
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