Question

Let $A$ be the set of points in the $XY$-plane which are equidistant from $P(-1, 0)$ and $Q(1, 0)$. Let $B$ be the set of points in the $XY$-plane which are equidistant from $A$ and $Q$. If $(5, y)$ is a point in $B$, then what is the value of $y^2$?

• A. 9
• B. 1
• C. 4
• D. 16

Hint:

The definition of set \(B\) is equivalent to the definition of a parabola.
The line \(x = 0\) acts as the directrix, and the point \(Q(1, 0)\) acts as the focus.
Remember that the standard equation of a parabola with focus \((a, 0)\) and directrix \(x = -a\) is \(y^2 = 4ax\).
Shifting the coordinates appropriately gives \(y^2 = 2x - 1\) directly.

Answer 1

The Correct Option is A

To solve this problem, we need to find two sets of loci and determine the intersection for a specific point.

First, we identify set \(A\), the set of points equidistant from \(P(-1, 0)\) and \(Q(1, 0)\). The locus of points equidistant from two points is the perpendicular bisector of the segment joining the points. The mid-point of \(PQ\) is \((0, 0)\).

• The slope of \(PQ\) is \(\frac{0 - 0}{1 - (-1)} = 0\). Therefore, the slope of the perpendicular bisector is undefined, indicating a vertical line.
• The vertical line through the midpoint is \(x = 0\).

Next, we examine set \(B\), the set of points equidistant from the line \(A\) (x = 0) and point \(Q(1, 0)\). To find this, we equate the distances from an arbitrary point \((x, y)\) to these entities:

• Distance from \((x, y)\) to \(x = 0\) is \(|x|\)
• Distance from \((x, y)\) to \((1, 0)\) is \(\sqrt{(x-1)^2 + y^2}\)

Given that \((5, y)\) lies in \(B\), substitute \(x = 5\) in the above equation:

1. \(y^2 = 2(5) - 1 = 10 - 1 = 9\)

Therefore, the value of \(y^2\) is 9.