The largest value of $k$ for which the circle $x^2+y^2=k^2$ lies completely in the interior of the parabola…
The largest value of $k$ for which the circle $x^2+y^2=k^2$ lies completely in the interior of the parabola $y^2=4 x+16$ is
- $4 \sqrt{3}$
- $2 \sqrt{3}$
- $2 \sqrt{6}$
- $4 \sqrt{6}$
Solution
Given, parabola is $y^2=4 x+16$
$
y^2=4(x+4)
$
Let parametric point on parabola be $\left(t^2-4,2 t\right)$ distance $O P \leq k$
$
\begin{array}{rlrl}
& & \left(t^2-4\right)^2+4 t^2 & \leq k^2 \\
t^4+16-8 t^2+4 t^2 & \leq k^2 \\
& & \\
t^4-4 t^2+16-k^2 & \leq 0 \\
& & D & \leq 0 \\
\Rightarrow & & 16-4\left(16-k^2\right) & \leq 0 \\
\Rightarrow & & 4-16+k^2 & \leq 0 \Rightarrow k^2 \leq 12 \\
k & \in[-2 \sqrt{3}, 2 \sqrt{3}]
\end{array}
$
Hence, option (2) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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