The normal at (2,32) to the ellipse x^216+y^23=1 touches a parabola, whose equation is:

Question

The normal at $\left(2,	frac32ight)$ to the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{3}=1$ touches a parabola, whose equation is:

Accepted Answer

Normal to $\dfrac{x^2}{16}+\dfrac{y^2}{3}=1$ at $(2,3/2)$: $\dfrac{16x}{2}-\dfrac{3y}{3/2}=16-3\Rightarrow8x-2y=13\Rightarrow y=4x-\dfrac{13}{2}.$

For $y=mx+c$ to be tangent to $y^2=4ax$: $c=\dfrac{a}{m}\Rightarrow a=mc=4\left(-\dfrac{13}{2}\right)=-26.$

Parabola: $y^2=4(-26)x=-104x.$

Solutions · 1