How to find $\angle$ b?

Question

How to find $\angle$ b ?

The vertices of the triangle are on the foci of the ellipse and on the ellipse.

$\angle$ a, the major axis and eccentricity are known.

Sorry about that, but apparently you cant post an image if rep is under 10. But it seems somebody can edit it in. — Andreas, Feb 24 '16 at 22:10
I was able to see it before Taussig edited....but I didn't know the answer:) — imranfat, Feb 24 '16 at 22:13

Nikunj · Answer 1 · 2016-02-25T09:22:01.017

Let the major axis be of length = $2a$, and the minor be of $2b$, then we know that $$b^2=a^2(1-e^2)$$ Foci are given by $P(ae,0)$,$Q(-ae,0)$ Let the point not on the vertex be $R(a\cos \theta,b\sin \theta)$ And I take angles RQP and RPQ as $\alpha$ and $\beta$ to avoid confusion.

Now we know that $$PQ=RQ \cos \alpha+RP \cos \beta$$ RQ and RP can be evaluated using the distance formula, We get $$RP=|a-ae\cos \theta| , RQ=|a+ae\cos \theta|$$ So now you say you know angle $a$ (i.e $\alpha$), also note that $a\cos \theta$ represents the x coordinate of the point R (not on the Foci) now we have all three sides of triangle PQR, applying the sine law in triangle PQR, we get

$$\frac{\sin \alpha}{RP}=\frac{\sin \beta}{RQ}$$ $$\implies \frac{\sin \alpha}{|a-ae\cos \theta|}=\frac{\sin \beta}{|a+ae\cos \theta|}$$ Now point R is also unknown to us, we can find it by considering the intersection of line QR with our ellipse, i.e R will be point of intersection of: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and $$y=\tan\alpha (x+ae)$$ Solving this, you would get the coordinates of R. As we only need the x coordinate, we can put that as $a\cos \theta$ and then find angle $\beta$ by the sine law in triangle PQR

Edit: Fixed RP and RQ

Are you sure that RP and RQ are correct? I also can't see how to get rid of $\theta$ as it's an unknown variable in this problem. I am not going to downvote yet, but your answer is not useful to solving this problem. — Andreas, Feb 24 '16 at 23:13
You don't have to think much about $\theta$, as you can replace the $x$ coordinate of the point R by $a\cos \theta$ directly, and yes RP,RQ are correct, use the distance formula and some manipulation. — Nikunj, Feb 25 '16 at 08:10
I am very certain that $RP=|1-ae\cos \theta| , RQ=|1+ae\cos \theta|$ should instead be $RP=|a-ae\cos \theta| , RQ=|a+ae\cos \theta|$ — Andreas, Feb 25 '16 at 09:16

Hugo · Answer 2 · 2016-02-25T11:26:15.633

Let $\alpha, \beta$ be the angles, $2a>0$ the length of the major axis, $P = (-ae, 0)$, $Q = (ae, 0)$ the foci, and $R = (x,y)$ the point on the ellipse. I assume for simplicity that $-ae < x < ae$ and $y >0$, since the calculations are analogous in other cases.

Note that $x,y$ are such that \begin{align} \frac{x^2}{a^2} + \frac{y^2}{(1-e^2)a^2} = 1 \end{align} and \begin{align*} y = \tan \alpha (x+ae). \end{align*}

We have \begin{align*} \tan \beta = \frac{y}{ae - x} = \frac{ae + x}{ae - x}\tan \alpha. \end{align*}

Let $t = (ae+x)/(ae-x)$, so that $x = ae(t-1)/(t+1)$ and $y = ae\tan \alpha (2t)/(t+1)$. Then the ellipse equation for $t$ is \begin{align*} e^2\frac{(t-1)^2}{(t+1)^2} + 4\tan^2 \alpha \frac{e^2}{1-e^2}\frac{t^2}{(1+t)^2} = 1. \end{align*}

If my calculations are correct, from this one gets \begin{align*} t = \frac{1-e^2}{4e^2\tan^2 \alpha - (1-e^2)^2}\left ( 1 + e^2 + \frac{2e}{\cos \alpha} \right ). \end{align*}

This times $\tan \alpha$ gives $\tan \beta$. Of course the value of $\alpha$ such that $2e\tan \alpha = 1-e^2$ is the one for which $x = ae$, since $t \to \infty$ gives $x \to ae$. Note also that for $\tan^2 \alpha = (1-e^2)/e^2$, i.e. when $R = (0, b)$, the expression simplifies to $t = 1$, as expected.

Unfortunately I cannot write the solution in a better form than this... Hope it can help anyway!

How to find $\angle$ b?

2 Answers2