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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

  1. A \(\frac{7 \sqrt{2}}{2}\)
  2. B \(\frac{7 \sqrt{2}}{16}\)
  3. C \(\frac{7 \sqrt{2}}{4}\)
  4. D \(\frac{7 \sqrt{2}}{8}\)
Verified Solution

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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