Generators of hyperbola

Question

A variable generator meets two generators of the system through extremities $B$ & $B'$ of the minor axis of the principal elliptic section of the hyperboloid $$\frac{x^2} {a^2} +\frac{y^2}{b^2} -z^2c^2=1$$ in $P$ & $P'$. Prove that $BP$. $B'P'=a^2+c^2$

My attempt : The point of intersection of p-sytem of generator with q-system of generator is $$x=\frac{a(1+pq)}{p+q},\ y=\frac{b(p-q)}{p+q},\ z=\frac{c(1-pq)}{p+q}$$ for the hyperbola $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2} {c^2} =1$. So, for the given hyperbola, $z=\frac{1-pq}{c(p+q)}$.

At $B(0,b,0)$ & $B'(0, - b, 0)$, $x=0$, $z=0$. So, $1+pq=0$, $1-pq=0$.

Not able to get $2$ values of $p$ or $q$.

Answer 1

We have the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2} {c^2} =1$

Generating lines of hyperboloid can be written in the standard form:

$$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}{-bcos\theta} = \frac{z}{c} \longrightarrow (1)$$ and $$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}{-bcos\theta} = \frac{z}{-c} \longrightarrow (2)$$

From (1), we can obtain the equations of two generators passing through minor axis by putting $\theta=90^\circ$ and $\theta=-90^\circ$: $$\frac{x}{a} = \frac{y - b}{0} = \frac{z}{c} \longrightarrow (3)$$ and $$\frac{x}{-a} = \frac{y + b}{0} = \frac{z}{c} \longrightarrow (4)$$

According to the question $B=(0,b,0)$ and $B'=(0,-b,0)$. Further P is the intersection of (2) and (3): $$P = (\frac{acos\theta}{1+sin\theta},b,\frac{cos\theta}{c(1+sin\theta)})$$ and P' is intersection of (2) and (4): $$P' = (\frac{-acos\theta}{sin\theta-1},-b,\frac{cos\theta}{c(sin\theta-1)})$$

Now we just apply distance formula $BP.B'P'$ and we get this as $a^2+c^2$. Hence proved.