# Focal length of an ellipse and related results

There are 2 questions(part of same question but I divided it into two):

Q1. Prove that the length of the focal chord of the ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ which is inclined to the major axis at an angle $\theta$ is $$\frac {2ab^2}{a^2\sin^2\theta+b^2\cos^2\theta}$$ I tried to solve this using the parametric form of a line, i.e., $(x,y)=(ae+r\cos\theta,r\sin\theta)$, plugging this into the given equation to find $r_1-r_2$ which is giving a different solution.

Q2. If $PSQ$ and $PS'R$ are two chords of an ellipse through its two focii S and S', then prove that $$\frac {PS}{SQ}+\frac {PS'}{S'R}=2\left(\frac {1+e^2}{1-e^2}\right)$$ Here I fixed $P \equiv (a\cos\theta,b\sin\theta)$ and found out the parametric angles of $Q$ and $R$ using the result $$\tan\theta_1/2*\tan\theta_2/2=\left(\frac {e-1}{e+1}\right)^{\pm1}$$, $+$ when chord passes through $S$ and $-$ when chord passes through $S'$. But it gives a very ugly expression. So how do I do it in an easy way? Hints are welcome.

• what is a focal chord? – abel Jan 2 '15 at 16:44
• a chord passing through the focus of the ellipse:- for a standard ellipse the focii are $(ae,0)$ and $(-ae,0)$ where e is the eccentricity.. – Abhishek Bakshi Jan 2 '15 at 16:46

Part I:

After parametrization $P(t)\equiv(ae+r\cos t,r\sin t)$: $$\frac{(ae+r\cos t)^2}{a^2}+\frac{r^2\sin^2t}{b^2}=1$$ $$b^2a^2e^2+b^2r^2\cos^2t+2ab^2er\cos t+a^2r^2\sin^2t-a^2b^2=0\\ (a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r+(b^2a^2e^2-a^2b^2)=0\\ (a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0\quad(\color{red}{b^2=a^2(1-e^2)})$$ Now: $$r_1=l_1,r_2=-l_2\implies |l_1+l_2|=|r_1-r_2|=\left|\sqrt{(r_1+r_2)^2-4r_1r_2}\right|$$ $$|l_1+l_2| =\sqrt{\left(\frac{2abe\cos t}{(a^2\sin^2t+b^2\cos^2t)}\right)^2+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\=\sqrt{\frac{4a^2b^4e^2\cos^2t}{(a^2\sin^2t+b^2\cos^2t)^2}+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\ =\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+\underbrace{b^2}_{b^2=a^2(1-e^2)}\cos^2t}\\ =\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+a^2\cos^2t-a^2e^2cos^2t}\\ \fbox{\huge PQ=\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}$$

Part II

$$(a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0$$

Now: $$\frac{PS}{SQ}=\frac{l_1}{l_2}=\left|\frac{r_1}{r_2}\right|=\left|\frac{(r_1+r_2)+(r_1-r_2)}{(r_1+r_2)-(r_1-r_2)}\right|=\left|\frac{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}+\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}-\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}\right|=\frac{1-e\cos t}{1+e\cos t}$$ Similiarly: $$\frac{PS'}{S'R}=\frac{1+e\cos u}{1-e\cos u}$$ Now take a triangle with vertices $P,S$ and the foot of perpendicular from P to Major Axis, then: $$\cos t=\frac{a\cos\theta-ae}{\sqrt{(a\cos\theta-ae)^2+b^2\sin^2\theta}}=\frac{\cos\theta-e}{1-e\cos\theta}$$ Similiarly: $$\cos u=\frac{\cos\theta+e}{1+e\cos\theta}$$ Now: $$\frac{PS}{SQ}=\frac{1-e\cos t}{1+e\cos t}=\frac{1-e\cos\theta-e(\cos\theta-e)}{1-e\cos\theta+e(\cos\theta-e)}=\frac{1-2e\cos\theta+e^2}{1-e^2}$$ Similiarly: $$\frac{PS'}{S'R}=\frac{1+2e\cos\theta+e^2}{1-e^2}$$ So: $$\frac{PS}{SQ}+\frac{PS'}{S'R}=\frac{1-2e\cos\theta+e^2}{1-e^2}+\frac{1+2e\cos\theta+e^2}{1-e^2}\\\fbox{\huge\frac{PS}{SQ}+\frac{PS'}{S'R}=2\left(\frac{1+e^2}{1-e^2}\right)}$$

• Are $\theta_1$ and $\theta_2$ the parametric angles of the ends of the focal chord? – Abhishek Bakshi Jan 2 '15 at 16:50
• thanks...you must have substituted $b^2=a^2(1-e^2)$...but in the first question we don't arrive at the given result...it comes something else.. – Abhishek Bakshi Jan 2 '15 at 16:59
• that was the exact expression that I got... – Abhishek Bakshi Jan 2 '15 at 17:11
• – Abhishek Bakshi Jan 2 '15 at 17:38
• @AbhishekBakshi was a typo, has been corrected – RE60K Jan 9 '15 at 8:34

First part

Using Newton polar form focal ray length of radius vector, total length is much easier to calculate.

$$=\frac{p}{1- e \,cos \theta }+ \frac{p}{1+ e \,cos \theta },\, e^2 = 1- (b/a)^2, p = b^2/a$$

and simplify to obtain the required result.

Second part

...also along similar lines with quotients instead of focal ray part sum.

• well, its quite late but thanks for the effort.. – Abhishek Bakshi Jan 11 '16 at 9:59