Point on a parabola at a given distance

Statement

Find the point $Q$ on the parabola $y=4-x^2$ that lies at distance $8$ from the point $P(6;2)$.

Solution

The unknowns are the coordinates of $Q=(x_Q;\,y_Q)$: two unknowns require two equations. The first translates the distance condition $PQ=8$, the second the fact that $Q$ lies on the parabola: $\begin{cases} (x_Q-6)^2+(y_Q-2)^2 = 64 \quad\text{(distanza} = 8) \\ y_Q = 4-x_Q^2 \quad\text{($Q$ sulla parabola)} \end{cases}$ We substitute the second equation into the first, obtaining a single equation in $x_Q$ alone: $(x_Q-6)^2+\bigl((4-x_Q^2)-2\bigr)^2 = 64,$ that is, a quartic equation in $x_Q$, whose real solutions give the abscissae of the sought points (and, through $y_Q=4-x_Q^2$, the ordinates).

Links

Topics: Euclidean circle
Concepts: Point–line distance · Line
Skills: Analytic geometry · Solving systems
Exercise type: Geometric problem