Parabola with a given vertex tangent to a line

Text

Among the infinitely many parabolas with vertex $V(-3;0)$, find the one(s) tangent to the line $y=-x+1$.

Solution

Every parabola with vertex $V(-3;0)$ has equation $y=a(x+3)^2$. Impose the intersection with the line: $a(x+3)^2 = -x+1 \implies a(x+3)^2 + x - 1 = 0.$ Setting $u=x+3$ (so $x=u-3$): $au^2 + (u-3) - 1 = au^2 + u - 4 = 0$. Tangency requires $\Delta = 1 + 16a = 0 \implies a = -\frac{1}{16}.$ There is a unique parabola: $\boxed{y=-\frac{1}{16}(x+3)^2}$ The point of tangency is $(5;-4)$.