#
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 \((8,0)\).