# Problem of the Week Problem E and Solution Sliding Parabola

## Problem

Suppose the parabola with equation $$y = 4 - x^2$$ has vertex at $$P$$ and crosses the $$x$$-axis at points $$A$$ and $$B$$, with $$B$$ lying to the right of $$A$$ on the $$x$$-axis.

This parabola is translated so that its vertex moves along the line $$y = x+ 4$$ to the point $$Q$$. The new parabola crosses the $$x$$-axis at points $$B$$ and $$C$$, with $$C$$ lying to the right of $$B$$ on the $$x$$-axis.

Determine the coordinates of $$C$$.

## Solution

For the original parabola $$y=-x^2+4$$, the vertex is $$P(0,4)$$ and the $$x$$-intercepts are $$A(-2,0)$$ and $$B(2,0)$$.

Let the vertex of the translated parabola be $$Q(q,p)$$. Since the new parabola is a translation of the original, the equation of this new parabola is $$y = -(x-q)^2 + p$$.

Since $$Q$$ lies on the line $$y=x+4$$, we have $$p = q + 4$$ and the equation of the new parabola is $$y=-(x-q)^2+q+4$$.

Since $$B(2,0)$$ lies on the new parabola, we can substitute $$(2,0)$$ into this equation: \begin{aligned} 0&=-(2-q)^2+q+4\\ 0&=-(q^2 -4q + 4) +q+ 4 \\ 0&=-q^2 + 5q\\ 0&=-q(q - 5) \end{aligned} Therefore, $$q=0$$ or $$q=5$$. The value $$q=0$$ corresponds to point $$P(0,4)$$ in the original parabola. Therefore, $$q=5$$. From here we will show two solutions.

Solution 1

Since $$q=5$$, the axis of symmetry for the new parabola is $$x=5$$. To find $$C$$ we need to reflect the point $$B(2,0)$$ in the axis of symmetry to get $$C(8,0)$$.

Solution 2

Since $$q=5$$, then the vertex of the new parabola is $$(5,9)$$ and the equation of this parabola is $$y=-(x-5)^2+9$$.

Since $$C$$ is an $$x$$-intercept of this parabola, to determine $$C$$ we set $$y=0$$ in the equation for the parabola and solve for $$x$$. \begin{aligned} 0&=-(x-5)^2+9\\ (x-5)^2&=9\\ x-5&=\pm 3\\ x&=8,2 \end{aligned} The value $$x=2$$ corresponds to point $$B$$, and the value $$x=8$$ corresponds to point $$C$$. Therefore, the coordinates of $$C$$ are $$(8,0)$$.