Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is…

Question

Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\dfrac12$. If $P(1,\beta)$, $\beta>0$ is a point on this ellipse, then the equation of the normal to it at $P$ is:

Accepted Answer

$\dfrac ae=4\Rightarrow a=2$; $b^2=a^2(1-e^2)=4\cdot\dfrac34=3$.

Ellipse: $\dfrac{x^2}4+\dfrac{y^2}3=1$. At $P(1,\beta)$: $\dfrac14+\dfrac{\beta^2}3=1\Rightarrow\beta=\dfrac32$.

Normal: $\dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2\Rightarrow\dfrac{4x}1-\dfrac{3y}{3/2}=1\Rightarrow4x-2y=1.$

Solutions · 1