A parabola has equation $y=k^{2}-x^{2}$, for some positive number $k$. Rectangle $A B C D$ is drawn with sides parallel to the axes so that $A$ and $D$ are the points where the parabola intersects the $x$-axis and so that the vertex, $V$, of the parabola is the midpoint of $B C$. If the perimeter of the rectangle is 48 , what is the value of $k$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-3.jpg?height=464&width=635&top_left_y=657&top_left_x=802)
Since the parabola has equation $y=k^{2}-x^{2}$, then its $y$-intercept is $k^{2}$, and so the height of rectangle $A B C D$ is $k^{2}$.

Since the parabola has equation $y=k^{2}-x^{2}=(k-x)(k+x)$, then its $x$-intercepts are $-k$ and $k$, so the width of rectangle is $k-(-k)=2 k$.

Since the perimeter is 48 , then $2 k^{2}+2(2 k)=48$ or $k^{2}+2 k-24=0$.

Factoring, we obtain $(k+6)(k-4)=0$.

Since $k>0$, then $k=4$.

ANSWER: 4