# Eliminate $\theta$ from from $\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$

I am stuck with the following problem :

Eliminate $$\theta$$ from $$\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$$ to get the following result:

$$k^2(c^2-z^2)^2=c^2\left(\,(cx \tan \alpha-yz)^2+(cy \cot \alpha-zx)^2\,\right)$$

I will be grateful if someone explains the problem which was actually the part of another problem . Thanks in advance for your time.

• Where is $y$ in the first equation? – Andrei May 21 at 18:41
• @Andrei sorry for the typo. corrected.. – learner May 21 at 18:44

Write your line as $$\frac{x-k\sin\theta\cot\alpha}{k\cos\theta}=\frac{z}{c}\quad\text{and}\quad\frac{y-k\cos\theta\tan\alpha}{k\sin\theta}=\frac{z}{c}.$$ Manipulate the two equations into \begin{align*} cx/k&=c\cot\alpha\sin\theta+z\cos\theta\\ cy/k&=z\sin\theta+c\tan\alpha\cos\theta \end{align*} so solving for $$\sin\theta,\cos\theta$$: \begin{align*} k(c^2-z^2)\sin\theta &=c\tan\alpha\cdot cx-z\cdot cy\\ k(c^2-z^2)\cos\theta &=c\cot\alpha\cdot cy-z\cdot cx \end{align*} Now remember $$\sin^2\theta+\cos^2\theta=1$$.
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