Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
"The" normal? What normal? At what point on the ellipse? –  DonAntonio Feb 16 '13 at 14:24
    
I suppose that he wants normals at all points on ellipse and then to choose the one which is more distant from center of the ellipse. –  zaarcis Feb 16 '13 at 14:26
    
Yes DonAntonio and point on ellipse –  juantheron Feb 16 '13 at 14:26
    
As zaarcis suggests, the question here is probably to compute the distance between the center and each normal, and then find the maximum of all these distances. For the first step, see here mathworld.wolfram.com/Point-LineDistance2-Dimensional.html. –  1015 Feb 16 '13 at 14:35
1  
@juantheron The maximum distance between a point and the points on a line is always $+\infty$. When we talk about the distance between a point $p$ and a line $L$, it means the minimum of $\|p-q\|$ where $q$ runs over all points of $L$. –  1015 Feb 16 '13 at 14:52

1 Answer 1

up vote 1 down vote accepted

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 $a>0,b>0$, $\tan^2\theta=\frac ba$

Now, $\frac{d^2f}{d\theta^2}=2a^2\left(\frac1{\cos^2\theta}+\frac{3\sin^2\theta}{\cos^4\theta}\right)+2b^2\left(\frac1{\sin^2\theta}+\frac{3\cos^2\theta}{\sin^2\theta}\right)>0$ 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.

share|improve this answer
    
Thanks lab bhattacharjee but This is a perpendicular Distance which is always Minimum So How can i Maximise that. –  juantheron Feb 16 '13 at 15:11
1  
@juantheron, distance implicitly means the perpendicular one. This answer calculates which normal has the maximum (perpendicular) distance from the origin. –  lab bhattacharjee Feb 16 '13 at 15:13
    
So the maximum distance is $\dfrac{a-b}{a+b}$ –  zaarcis Feb 16 '13 at 15:41
1  
@zaarcis, the maximum distance will be $\left|\frac{a^2-b^2}{a+b} \right|=\mid a-b\mid$ –  lab bhattacharjee Feb 16 '13 at 15:45
    
My fault. Forget about square root. –  zaarcis Feb 16 '13 at 15:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.