Test for [math]\theta=20[/math] and [math]\phi=0[/math]
Solving for the components of the ellipse
[math]a_1=\frac{2.53sin(\theta)}{sin(115-\theta)}=\frac{2.53sin(20^{\circ})}{sin(115^{\circ}-20^{\circ})}=\frac{2.53\cdot.342}{.9962}=.8686[/math]
[math]a_2=\frac{2.53sin(\theta)}{sin(65-\theta)}=\frac{2.53sin(20^{\circ})}{sin(65^{\circ}-20^{\circ})}=\frac{2.53\cdot.342}{.0872}=1.2237[/math]
[math]c_{ellipse} \equiv (-\Delta a\ sin(65^{\circ}), 0,z+\Delta a\ cos(65^{\circ}))[/math]
[math]c_{ellipse} \equiv (-\frac{a_2-a_1}{2} sin(65^{\circ}), 0,2.53+\frac{a_2-a_1}{2} cos(65^{\circ}))[/math]
[math]c_{ellipse} \equiv (-\frac{1.2237-.8686}{2} sin(65^{\circ}), 0,2.53+\frac{1.2237-.8686}{2} cos(65^{\circ}))[/math]
[math]c_{ellipse} \equiv (-\frac{.3550}{2} sin(65^{\circ}), 0,2.53+\frac{.3550}{2} cos(65^{\circ}))[/math]
[math]c_{ellipse} \equiv (-.1775 sin(65^{\circ}), 0,2.53+.1775 cos(65^{\circ}))[/math]
[math]c_{ellipse} \equiv (-.1609, 0,.0750)[/math]
[math]P(\phi=0) \equiv (-\frac{a_2-a_1}{2} sin(65^{\circ})+\frac{a_1+a_2}{2} sin((65^{\circ}), 0,2.53+\frac{a_2-a_1}{2} cos(65^{\circ})-\frac{a_1+a_2}{2} cos(65^{\circ}))[/math]
[math]P(\phi=0) \equiv (-\frac{1.2234-.8684}{2} sin(65^{\circ})+\frac{.8684+1.2234}{2} sin((65^{\circ}), 0,2.53+\frac{1.2234-.8684}{2} cos(65^{\circ})-\frac{.8684+1.2234}{2} cos(65^{\circ}))[/math]
[math]P(\phi=0) \equiv (-\frac{.3550}{2} sin(65^{\circ})+\frac{2.0918}{2} sin((65^{\circ}), 0,2.53+\frac{.3550}{2} cos(65^{\circ})-\frac{2.0918}{2} cos(65^{\circ}))[/math]
[math]P(\phi=0) \equiv (-.1775 sin(65^{\circ})+1.0459 sin((65^{\circ}), 0,2.53+.1775 cos(65^{\circ})-1.0459 cos(65^{\circ}))[/math]
[math]P(\phi=0) \equiv (.7870, 0,2.1623)[/math]
[math]x=\frac{2.53cos(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}[/math]
[math]x=\frac{2.53cos(0)}{cot(20^{\circ})+cos(0)cot(65^{\circ})}[/math]
[math]x=\frac{2.53}{2.7475+.4663}[/math]
[math]x=\frac{2.53}{3.2138}=.7870\text{cm}[/math]
The y component is zero for [math]\phi=0[/math]
The z component can be found from the ellipse equation
[math]z=-cot(65^{\circ})x+2.53=-cot(65^{\circ})(.7870)+2.53=2.1623\ \text{cm}[/math]
[math]e\equiv \frac{sin(25^{\circ})}{cos(20^{\circ})}=0.96447[/math]
[math]r_{D1}=R_{Lower\ Dandelin}cos(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=R_{Lower\ Dandelin}cos(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)[/math]
[math]r_{D1}=(\frac{a_1+a_2}{2}e-\frac{a_2-a_1}{2}) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=(\frac{a_1+a_2}{2}e+\frac{a_2-a_1}{2}) tan(65^{\circ})cos(\theta)[/math]
[math]r_{D1}=(\frac{.8686+1.2237}{2}e-\frac{1.2237-.8686}{2}) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=(\frac{.8686+1.2237}{2}e+\frac{1.2237-.8686}{2}) tan(65^{\circ})cos(\theta)[/math]
[math]r_{D1}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}-\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ})\qquad \qquad r_{D2}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}+\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ}))[/math]
[math]r_{D1}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}-\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ})\qquad \qquad r_{D2}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}+\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ}))[/math]
[math]r_{D1}=(1.0459(.4497)-.1775)\cdot 2.1445\cdot .9397=.5901\ \text{m} \qquad \qquad r_{D2}=(1.0459(.4497)+.1775)\cdot 2.1445\cdot .9397=1.3055 \text{m}[/math]
The height to the first directrix circle is
[math]z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212[/math]
and the x and y components are
[math]x_{D1}=r_{D1}\ cos(\phi)=.5901 cos(0)=.5901\text {m}\ \ \ \ y_{D1}=r_{D1}cos(\phi)=.5901 sin(0)=0[/math]
The height to the second directrix circle is
[math]z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m}[/math]
and the x and y components are
[math]x_{D2}=r_{D2} cos(\phi)=1.3055cos(0)=1.3055\text {m}\ \ \ \ y_{D2}=r_{D2} sin(\phi)=1.3055sin(0)=0[/math]
The distance between the two point
[math]\sqrt{(1.3055-.5901)^2+(0-0)^2+(3.5868-1.6212)^2}=\sqrt{.715^2+1.9656^2}=2.09\text {m}[/math]
This should be equal to 2a
[math]2a=2(\frac{a_1+a_2}{2})=2(\frac{.8684+1.2234}{2})=2.09\text {m}[/math]
This gives the components for a point on the directrix circles as
| [math]x_{D1}=r_{D1}\ cos(\phi)=.5901cos(0)=.5901\text {m}\qquad y_{D1}=r_{D1}cos(\phi)=.5901 sin(0)=0\qquad z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212\ \text{m}[/math]
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| [math]x_{D2}=r_{D2} cos(\phi)=1.3055cos(0)=1.3055\text {m}\qquad y_{D2}=r_{D2} sin(\phi)=1.3055sin(0)=0\qquad z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m}[/math]
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| [math]x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cos(0)}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=0.7872\qquad y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53sin(0)}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=0\qquad z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cot(20^{\circ})}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=2.1623[/math]
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[math]D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}=\sqrt{(1.3055-0.7872)^2+(0-0)^2+(3.5868-2.1623)^2}=\sqrt{(.5183)^2+(1.445)^2}=\sqrt{.2686+2.0292}=1.5158611\ \text{m}[/math]
[math]D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7872-.5901)^2+(0-0)^2+(2.1623-1.6212)^2}=\sqrt{(.1971)^2+(.5411)^2}=\sqrt{.0388+.2927}=.575879865944\ \text{m}[/math]