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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

juantheron
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  • "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. – zaa 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 http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html. – Julien Feb 16 '13 at 14:35
  • To julien I have one Question. Distance of point $P(x_{1},y_{1})$ to line $ax+by+c = 0$ is $=\displaystyle \left|\frac{ax_{1}+by_{1}+c}{a^2+b^2}\right|$ which is a Minimum distance but How can I calculate for Maximum – juantheron Feb 16 '13 at 14:37
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    @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 $L$. – Julien Feb 16 '13 at 14:52

2 Answers2

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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.

lab bhattacharjee
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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 <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

Chotu
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