# Trigonometry and algebra question

Given:

• The total length of ad + dc
• The lengths of each ab, bc and ca
• That angle adb = angle cdb

How can I work out lengths ad and cd

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Can you either give a drawing or a description of it? Here we don't know what a, b, c, d are expected to be. – Jean-Claude Arbaut Jun 28 '13 at 7:23
There should be an image in the question. I can see it, can anyone else? Try this link – Jake Jun 28 '13 at 7:45
Ok, I can see it now. – Jean-Claude Arbaut Jun 28 '13 at 10:07

Geometric Solution:

Let $h=\dfrac{|\overline{bc}|}2$ and $k=\dfrac{|\overline{bd}|+|\overline{cd}|}2$.

Place $c$ at $(-h,0)$ and $b$ at $(+h,0)$. $a$ would then be at the intersection of the circle of radius $|\overline{ac}|$ centered at $c$ and the circle of radius $|\overline{ab}|$ centered at $b$.

Since $|\overline{bd}|+|\overline{cd}|=2k$ (i.e. a constant), $d$ lies on the ellipse $$\frac{x^2}{k^2}+\frac{y^2}{k^2-h^2}=1\tag{1}$$ Graphically,

$\hspace{3.2cm}$

Because of the reflection property of ellipses, $\angle adc=\angle adb$ when $\overline{ad}$ is normal to the ellipse (then $\angle adc$ and $\angle adb$ are supplementary to the angle of incidence). That is, $d$ is the point on the ellipse given in $(1)$ closest to (or furthest from) $a$.

The slope of the normal to the ellipse at $d=(x,y)$ is $$\frac yx\frac{k^2}{k^2-h^2}\tag{2}$$ Therefore, we need $$\frac{y-a_y}{x-a_x}=\frac yx\frac{k^2}{k^2-h^2}\tag{3}$$ which becomes the right hyperbola $$\left(x-\frac{k^2}{h^2}a_x\right)\left(y+\frac{k^2-h^2}{h^2}a_y\right) =-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{4}$$ Thus, we can find $d$ at the intersection of the ellipse in $(1)$ and the right hyperbola in $(4)$.

For computation of $d$, it is probably easiest to use $$u=x-\frac{k^2}{h^2}a_x\quad\text{and}\quad v=y+\frac{k^2-h^2}{h^2}a_y\tag{5}$$ Then, we have from $(1)$ and $(4)$, $$\frac{\left(u+\frac{k^2}{h^2}a_x\right)^2}{k^2}+\frac{\left(v-\frac{k^2-h^2}{h^2}a_y\right)^2}{k^2-h^2}=1\tag{6}$$ and $$uv=-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{7}$$ Using $(7)$ in $(6)$ yields a fourth degree equation to solve.

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Looks like you've put me on the right track. With h=3, k=4, ax=-2, ay=6 I get this graphically Now I just need to solve for the intersection. – Jake Jun 30 '13 at 3:57
Thanks for your help! I follow what you've done so far but I'm still unsure how to proceed. I've tried rearranging (7) for u and substituting the result into (6) but from there I struggle to rearrange to solve for v. I'm sure it would be simple if I could plug in k,h and a but I need to be able to use this for any (reasonable) values of those variables. – Jake Jul 1 '13 at 10:19
With the values you gave in your previous comment, $(7)$ becomes $$uv=\frac{448}{27}$$ and $(6)$ becomes $$\frac{\left(u-\frac{32}{9}\right)^2}{16}+\frac{\left(v-\frac{42}{9}\right)^2}{‌​7}=1$$ Combining these gives $$\frac{\left(u-\frac{32}{9}\right)^2}{16} +\frac{\left(\frac{448}{27u}-\frac{42}{9}\right)^2}{7}=1$$ which gives a quartic equation to solve for $u$. – robjohn Jul 1 '13 at 12:31
Sorry to keep pestering you, but I can't work this out. I need to be able write some code to solve for any values of k, h, and ax,ay (in response to a user dragging point a). Could you explain how to solve the equation above for u? The only methods I know for dealing with quadratics don't seem to be possible with this. – Jake Jul 3 '13 at 7:22
@Jake: If you multiply the last equation by $u^2$ and subtract off the $u^2$ on the right, you get a quartic equation: $f(u)=0$. To solve this, you can use Newton's Method: $$u_{n+1}=u_n-\frac{f(u_n)}{f'(u_n)}$$ – robjohn Jul 3 '13 at 13:30