If the tangents on the ellipse 4x^2+y^2=8 at the points (1,2) and (a,b) are perpendicular …

Question

If the tangents on the ellipse $4x^2+y^2=8$ at the points $(1,2)$ and $(a,b)$ are perpendicular to each other, then $a^2$ is equal to:

Accepted Answer

$\dfrac{x^2}2+\dfrac{y^2}8=1$. Tangent at $(1,2)$: $\dfrac x2+\dfrac{2y}8=1\Rightarrow$ slope $-2$.

Tangent at $(a,b)$ has slope $-\dfrac{4a}b$; perpendicularity: $\left(-\dfrac{4a}b\right)(-2)=-1\Rightarrow b=-8a$.

On the ellipse: $4a^2+64a^2=8\Rightarrow68a^2=8\Rightarrow a^2=\dfrac2{17}.$

Solutions · 1