If the maximum distance of a normal to the ellipse x^24+y^2b^2=1, b

Question

If the maximum distance of a normal to the ellipse $\dfrac{x^2}4+\dfrac{y^2}{b^2}=1$, $b<2$, from the origin is $1$, then the eccentricity of the ellipse is:

Accepted Answer

Normal at $P(2\cos\theta,b\sin\theta)$: $2x\sec\theta-by\csc\theta=4-b^2$. Distance from origin:
$$d=\frac{4-b^2}{\sqrt{4\sec^2\theta+b^2\csc^2\theta}}.$$

Minimizing $4\tan^2\theta+b^2\cot^2\theta\ge4b$ (AM--GM) gives $4\sec^2\theta+b^2\csc^2\theta\ge (b+2)^2$ at minimum, so
$$d_{\max}=\frac{(2-b)(2+b)}{2+b}=2-b=1\ \Rightarrow\ b=1.$$

$$e=\sqrt{1-\frac{b^2}{4}}=\sqrt{1-\frac14}=\frac{\sqrt3}2.$$

Solutions · 1