Let $L$ be the distance between two parallel normals of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$, then find maximum value of $L$

Question

Let the equation of normal be $$ax\sec t -by\csc t=a^2-b^2$$

Let the parameters of the two normals be $t$ and $r$

Their slopes are equal $$\frac{\sec t}{\csc t}=\frac{\sec r}{\csc r}$$

$$\sin t \cos r-\cos t \sin r=0$$

$$t-r=n\pi$$ Let $n=1$

$$t-r=\pi$$ $$t=\pi +r$$

So the first equation becomes $$ax\sec r -by\csc r =b^2-a^2$$

The distance between two parallel lines is $$\frac{|a^2-b^2+b^2+a^2|}{\sqrt{ a^2\sec ^2r+b^2\csc^2 r}}$$

To minimise the denominator , differentiating wrt $r$

$$2a^2\sec^2r\tan r-2b^2\csc^2r\cot r=0$$

$$\tan ^2r=\frac ab$$

Then $$L =\frac{2a^2}{a+b}$$

The given answer is $2(a-b)$, but I am unable to find what is wrong in my computation.

Thanks for the help

Answer 1

Note that your distance expression is off. Given symmetry, the distance between the two parallel normals is twice the distance from origin to either line. So, continue with $$ax\sec t -by\csc t=a^2-b^2$$ to express the distance as

$$d(t)=\frac{2(a^2-b^2)}{\sqrt{ a^2\sec ^2t+b^2\csc^2 t}}\tag1$$

Set $d’(t)=0$ to get $\tan t= \pm\sqrt{\frac ba }$ and substitute into (1) to obtain the maximum distance

$$d_{max}= \frac{2(a^2-b^2)}{\sqrt{ a^2+b^2+2ab}}= 2(a-b)$$

Answer 2

Rather than using parametric coordinates, one can express the normal to an ellipse in the slope for

For a given slope $m$, the normal to an ellipse can be expressed as

$$y = mx \pm \frac{m(a^2-b^2)}{\sqrt{a^2 + b^2m^2}}$$

Hence, for any slope, the distance can be expressed as

$$L(m) = \frac{d}{\sqrt{1+m^2}}$$

where $d = \frac{2m(a^2-b^2)}{\sqrt{a^2 + b^2m^2}}$

Hence

$$L(m) = (a^2-b^2). \frac{m}{\sqrt{1+m^2}\sqrt{a^2 + b^2m^2}}$$

Answer 3

Okay, another try.
Given $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and performing implicit differentiation we have $$\frac{2x}{a^2}+\frac{2y'y}{b^2}=0$$ I won't consider a slope 'cause it may be equal to $\pm\infty$, instead a normal vector will be $$\left(\frac{x}{a^2},\frac{y}{b^2}\right) \hbox{ or }(b^2x,a^2y)$$ with parametrization $x=a\cos t,\,y=a\sin t$ it will be $$(b\cos t,a\sin t)$$ Since the lines for parameters $t,r$ are collinear, we have cross product $=0$: $$(b\cos t)(a\sin r)-(b\cos r)(a\sin t)=0 \\ \sin(r-t)=0,\hbox{ so }\\r-t=\pi n$$ So the points will be $$(-a\cos r,-b\sin r),\,(a\cos r,b\sin r)$$ with the of difference $$(2a\cos r,2b\sin r)$$
and we have to dot multiply by the perpendicular to the normal vector and divide by its length to get the desired distance: $$\begin{align*}\frac{|2(a\cos r,b\sin r).(-a\sin r,b\cos r)|}{\sqrt{(b\cos r,a\sin r)^2}}&=\frac{(a^2-b^2)|\sin(2r)|}{\sqrt{b^2\cos^2 r+a^2(1-\cos^2 r)}}\\ &=\frac{(a^2-b^2)|\sin(2r)|}{\sqrt{(b^2-a^2)\cos^2 r+a^2}}\\ &=\frac{(a^2-b^2)|\sin(2r)|}{\sqrt{(b^2-a^2)\frac{\cos(2r)+1}{2}+a^2}} \end{align*}$$ so the formula for $L$ seems to correspond with OP's one.
Equating the derivative to $0$ we get $$2r=2\pi n-2\arctan\left(\frac{\sqrt{b}}{\sqrt{a}}\right)$$ and substituting it back we have $$\frac{2(a^2-b^2)}{a+b}$$ so I'd suggest expanding the lines after finding the explicit formula for $L(r)$ so we can see if is there a mistake. Thanks.