Question

If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP².

Answer 1

Equation of the hyperbola:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

If the axes of the hyperbola are equal, then $a=b$.
Then, equation of the hyperbola becomes $x^2-y^2=a^2$.

$$\begin{gathered} \therefore b^2=a^2\left(e^2-1\right) \\ \Rightarrow a^2=a^2\left(e^2-1\right) \\ \Rightarrow 1=\left(e^2-1\right) \\ \Rightarrow e^2=2 \\ \Rightarrow e=\sqrt{2} \end{gathered}$$
Thus, the centre $C\left(0,0\right)$ and the focus are given by $S\left(\sqrt{2}a,0\right)$ and $S\text{'}\left(-\sqrt{2}a,0\right)$, respectively.
Let $P\left(\alpha ,\beta \right)$ be any point on the parabola.
So, it will satisfy the equation.
${\alpha }^2-{\beta }^2=a^2$

$$\begin{gathered} \therefore S{P}^2={\left(\sqrt{2}a-\alpha \right)}^2+{\beta }^2 \\ =2a^2+{\alpha }^2-2\sqrt{2}a\alpha +{\beta }^2 \end{gathered}$$

$$\begin{gathered} S\text{'}{P}^2={\left(-\sqrt{2}a-\alpha \right)}^2+{\beta }^2 \\ =2a^2+{\alpha }^2+2\sqrt{2}a\alpha +{\beta }^2 \end{gathered}$$

$$\begin{gathered} Now,S{P}^2.S\text{'}{P}^2=\left(2a^2+{\alpha }^2-2\sqrt{2}a\alpha +{\beta }^2\right)\left(2a^2+{\alpha }^2+2\sqrt{2}a\alpha +{\beta }^2\right) \\ =4a^4+4a^2\left({\alpha }^2+{\beta }^2\right)+{\left({\alpha }^2+{\beta }^2\right)}^2-8a^2{\alpha }^2 \\ =4a^2\left(a^2-2{\alpha }^2\right)+4a^2\left({\alpha }^2+{\beta }^2\right)+{\left({\alpha }^2+{\beta }^2\right)}^2 \\ =4a^2\left({\alpha }^2-{\beta }^2-2{\alpha }^2\right)+4a^2\left({\alpha }^2+{\beta }^2\right)+{\left({\alpha }^2+{\beta }^2\right)}^2 \\ =-4a^2\left({\alpha }^2+{\beta }^2\right)+4a^2\left({\alpha }^2+{\beta }^2\right)+{\left({\alpha }^2+{\beta }^2\right)}^2 \\ ={\left({\alpha }^2+{\beta }^2\right)}^2 \\ =C{P}^4 \end{gathered}$$
∴ $SP.S\text{'}P=C{P}^2$