Points of a parabola at a fixed distance from a line

Text

Find the points of the parabola $y=4-x^2$ that are exactly at distance $1$ from the line $y=\dfrac{3x}{4}$.

Solution

The line is $3x-4y=0$, with $\sqrt{3^2+4^2}=5$. A point of the parabola is $P(t;\,4-t^2)$; impose distance $1$: $\frac{|3t-4(4-t^2)|}{5}=1 \implies |4t^2+3t-16|=5.$ Two equations follow:

• $4t^2+3t-21=0 \implies t=\dfrac{-3\pm\sqrt{345}}{8}$;
• $4t^2+3t-11=0 \implies t=\dfrac{-3\pm\sqrt{185}}{8}$.

This gives four points, with abscissas $t\approx 1{,}95,\ -2{,}70,\ 1{,}32,\ -2{,}08$ and ordinate $4-t^2$.

$\boxed{t=\frac{-3\pm\sqrt{345}}{8}\quad\text{and}\quad t=\frac{-3\pm\sqrt{185}}{8}}$