Max. distance of Normal to ellipse from origin How Can I calculate Maximum Distance of Center of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ from the Normal.

My Try ::  Let $P(a\cos \theta,b\sin \theta)$ be any point on the ellipse. Then equation of Normal at that point is

$ax\sec \theta-by\csc \theta = a^2-b^2$. Then How can I find Max. distance of Center of the ellipse from the Normal
So, the distance of the normal from the origin $(0,0)$ is $$\left| \frac{a^2-b^2}{\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}} \right|$$

So, we need to minimize $(a\sec\theta)^2+(-b\csc\theta)^2=a^2\sec^2\theta+b^2\csc^2\theta=f(\theta)$(say)

So, $\frac{df}{d\theta}=a^22\sec\theta\sec\theta\tan\theta+b^22\csc\theta(-\csc\theta\cot\theta)=2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}$

For the extreme value of $f(\theta),\frac{df}{d\theta}=0$

$\implies 2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}=0$ or $\tan^4\theta=\frac{b^2}{a^2}$ 

Assuming $a&gt;0,b&gt;0$, $\tan^2\theta=\frac ba$

Now, $\frac{d^2f}{d\theta^2}=2a^2\left(\frac1{\cos^2\theta}+\frac{3\sin^2\theta}{\cos^4\theta}\right)+2b^2\left(\frac1{\sin^2\theta}+\frac{3\cos^2\theta}{\sin^2\theta}\right)&gt;0$ for real $\theta$

So, $f(\theta)$ will attain the minimum value at $\tan^2\theta=\frac ba$

So, $f(\theta)_\text{min}=a^2\sec^2\theta+b^2\csc^2\theta_{\text{at  }\tan^2\theta=\frac ba}=a^2\left(1+\frac ba\right)+b^2\left(1+\frac ab\right)=(a+b)^2$

So, the minimum value of $\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}$ is $a+b$

If $\tan\theta=\sqrt \frac ba, \frac{\sin\theta}{\sqrt b}=\frac{\cos\theta}{\sqrt a}=\pm\frac1{b+a}$

If $\sin\theta=\frac{\sqrt b}{a+b}\implies \csc\theta=\frac{a+b}{\sqrt b},\cos\theta=\frac{\sqrt a}{a+b}\implies \sec\theta=\frac{a+b}{\sqrt a}$

There will be another set $(\csc\theta=-\frac{a+b}{\sqrt b},\sec\theta=-\frac{a+b}{\sqrt a})$

There will be two more set of values of $(\csc\theta,\sec\theta)$ for $\tan\theta=-\sqrt\frac ba$

So, we shall have four normals having the maximum distance from the origin.
let a point p(acost,bsint) is on the ellipse.
x2/a2 y2/b2=1
dy/dx=-b2x/a2y

dy/dx of normal on(acost,bsint)
= a2y/b2x=a/btant
equestion of normal
y-bsint=b/atant(x-acost)
axsect-bycost-(a2-b2)=0

now
lenth from origin(0,0) of the normal
l=mode -(a2-b2)/ squat(a2sec2t b2cosec2t)
=a2-b2/squat(a2sec2t b2cosec2t)
then
Differential of this function respect t
dl/dt=a2-b2(b2cos4t-a2sin4t)/(a2sin2t b2cosec2t)3/2
for max. lenth
dl/dt=0
b2cos4t-a2sin4t =0
tant=squat(b/a)
now
d2l/dt2=(a2-b2)(-4sintcost)[(a2sin2t b2cos2t)3/2]
now
 d2l/dt2 on tant=squat(b/a)                                                  ' .'sint=squat(b/a b),cost=squat(a/a b)
    =  -4(a-b)a2b2   &lt;0    ;.a>b 
so the lenth of normal will max. on t=tan(inverse)squat(b/a)
now
 l=a2-b2/squat(a2sec2t b2cosec2t)
 l(max.)=  a2-b2/squat{a2(a b/a) b2(a b?b)           ;.sect=squat{(a b)/a,},cosect=squat{(a b)/b}   because tant=squat(b/a)
             =  a2-b2/squat{(a b)(a b)}
              = a2-b2/a b
               =a-b