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(D) The given parabolas are symmetric about the line $y = x$.Tangents at points $A$ and $B$ must be parallel to the line $y = x$,so the slope of the t

Vedclass · Accepted Answer

$\frac{7 \sqrt{2}}{8}$

Vedclass · Answer

(D) The given parabolas are symmetric about the line $y = x$.Tangents at points $A$ and $B$ must be parallel to the line $y = x$,so the slope of the tangents is $1$.For the parabola $y = x^2 + 2$,we have $\frac{dy}{dx} = 2x = 1$,which gives $x = \frac{1}{2}$.Substituting $x = \frac{1}{2}$ into $y = x^2 + 2$,we get $y = (\frac{1}{2})^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4}$.Thus,point $B = (\frac{1}{2}, \frac{9}{4})$. By symmetry,point $A = (\frac{9}{4}, \frac{1}{2})$.The distance $AB$ is $\sqrt{(\frac{9}{4} - \frac{1}{2})^2 + (\frac{1}{2} - \frac{9}{4})^2} = \sqrt{(\frac{7}{4})^2 + (-\frac{7}{4})^2} = \sqrt{\frac{49}{16} + \frac{49}{16}} = \sqrt{\frac{98}{16}} = \frac{7 \sqrt{2}}{4}$.The diameter of the smallest circle is the distance $AB$,so the radius $r = \frac{AB}{2} = \frac{7 \sqrt{2}}{8}$.

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

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