Statement 1: The line x-2y=2 meets the parabola y^2+2x=0 only at the point (-2,-2).

Question

Statement 1: The line $x-2y=2$ meets the parabola $y^2+2x=0$ only at the point $(-2,-2)$.
Statement 2: The line $y=mx-\dfrac{1}{2m}\ (m
e0)$ is tangent to the parabola $y^2=-2x$ at the point $\left(-\dfrac{1}{2m^2},-\dfrac1might)$.

Accepted Answer

From $x-2y=2$, $x=2y+2$. Substituting into $y^2=-2x=-2(2y+2)$: $y^2+4y+4=0\Rightarrow(y+2)^2=0\Rightarrow y=-2$ (double root), $x=-2$. So the line meets the parabola only at $(-2,-2)$ --- tangentially --- confirming Statement 1.

Writing $x-2y=2$ as $y=	frac12x-1$, i.e. $m=	frac12$: Statement 2's formula gives the point of tangency $\left(-2,-2ight)$, matching exactly, and correctly explains why the line meets the parabola at just that one point. Statement 2 is true and is a correct explanation.

Solutions · 1