A point on the ellipse 4x^2+9y^2=36, where the normal is parallel to the line 4x-2y-5=0, i…

Question

A point on the ellipse $4x^2+9y^2=36$, where the normal is parallel to the line $4x-2y-5=0$, is:

Accepted Answer

Ellipse: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, so $a^2=9,\ b^2=4$. Slope of normal $=2\Rightarrow$ slope of tangent $=-\tfrac12$.

Point of contact for slope $m$: $\left(\dfrac{-a^2m}{\sqrt{a^2m^2+b^2}},\dfrac{b^2}{\sqrt{a^2m^2+b^2}}\right)$. With $m=-\tfrac12$: $a^2m^2+b^2=9\cdot\tfrac14+4=\tfrac{25}{4}$, $\sqrt{\cdot}=\tfrac52$.
$$x=\frac{-9(-\tfrac12)}{5/2}=\frac{9/2}{5/2}=\frac95\cdot(-1)=-\frac95,\qquad y=\frac{4}{5/2}=\frac85.$$

Point: $\left(-\tfrac95,\tfrac85\right)$.

Solutions · 1