The largest value of k for which the circle x | vedclass

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Question

(B) The equation of the parabola is $y^2 = 4(x+4)$.Let a point on the parabola be $P(x, y) = (t^2-4, 2t)$.The distance squared from the origin $(0,0)$

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$2\sqrt{3}$

Vedclass · Answer

(B) The equation of the parabola is $y^2 = 4(x+4)$.Let a point on the parabola be $P(x, y) = (t^2-4, 2t)$.The distance squared from the origin $(0,0)$ to any point on the parabola is $d^2 = x^2 + y^2 = (t^2-4)^2 + (2t)^2$.$d^2 = t^4 - 8t^2 + 16 + 4t^2 = t^4 - 4t^2 + 16$.For the circle $x^2+y^2=k^2$ to lie inside the parabola,we must have $k^2 \leq d^2$ for all $t$.Let $u = t^2$,where $u \geq 0$. Then $f(u) = u^2 - 4u + 16$.The minimum value of $f(u)$ occurs at $u = -(-4)/(2 	imes 1) = 2$.Since $u=2 \geq 0$,the minimum value is $f(2) = 2^2 - 4(2) + 16 = 4 - 8 + 16 = 12$.Thus,$k^2 \leq 12$,which implies $k \leq \sqrt{12} = 2\sqrt{3}$.The largest value of $k$ is $2\sqrt{3}$.

The largest value of $k$ for which the circle $x^2+y^2=k^2$ lies completely in the interior of the parabola $y^2=4x+16$ is

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