Tangent lines to a parabola from an external point

Text

Given the parabola $y = 2 - 2x^2$, find the equations of the two tangent lines to its graph passing through the point $A(-2;2)$.

Solution

The point $A(-2;2)$ lies outside the parabola (indeed $y(-2)=2-8=-6\neq 2$). The pencil of lines through $A$ is $y = m(x+2) + 2.$ Impose the intersection with the parabola: $m(x+2)+2 = 2-2x^2 \implies 2x^2 + mx + 2m = 0.$ The tangency condition is a zero discriminant: $\Delta = m^2 - 4\cdot 2\cdot 2m = m^2 - 16m = 0 \implies m(m-16)=0 \implies m=0 \ \vee\ m=16.$

• $m=0$: line $y=2$ (horizontal tangent at the vertex $(0;2)$);
• $m=16$: line $y=16(x+2)+2 = 16x+34$ (tangent at $(-4;-30)$).

$\boxed{y=2 \qquad\text{and}\qquad y=16x+34}$