Question

If P is any point on the hyperbola whose axis are equal, prove that SP.SP=$C{P}^2$

Answer 1

$\text{Equation of the hyperbola:}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\text{If the axes of the hyperbola are equal,then a=b}$

$\text{Then equation of the hyperbola becomes}$

$x^2-y^2=a^2$ $\therefore b^2=a^2(e^2-1)$

$\Rightarrow a^2=a^2(e^2-1)$

$\Rightarrow 1=(e^2-1)$

$\Rightarrow e^2=2$

$\Rightarrow e=\sqrt{2}$

$\text{Thus,the centre C(0,0) and the focus are given by X}(\sqrt{2}a,0)and{S}^{\prime }\left)\right(-\sqrt{2}a,0),respectively.$

$LetP(\alpha ,\beta )\text{be any point on the parabola.}$

$\text{So,it will satisfy the equation.}$

${\alpha }^2-{\beta }^2=a^2$

$\therefore S{P}^2=(\sqrt{2}a-\alpha {)}^2+{\beta }^2$

$=2a^2+{\alpha }^2-2\sqrt{2}a\alpha +\beta$

$S^{\prime }{P}^2=(-\sqrt{2}a-\alpha {)}^2+{\beta }^2$

$=2a^2+{\alpha }^2+2\sqrt{2}a\alpha +\beta$

$Now,S{P}^2,S^{\prime }{P}^2=(2a^2+{\alpha }^2-2\sqrt{2}a\alpha +\beta )(2a^2+{\alpha }^2+2\sqrt{2}a\alpha +{\beta }^2)$

$=4a^4+4a^2({\alpha }^2+{\beta }^2)+({\alpha }^2+{\beta }^2{)}^2-8a^2{\alpha }^2$

$=4a^2(a^2-2{\alpha }^2)+4a^2({\alpha }^2+{\beta }^2)+({\alpha }^2+{\beta }^2{)}^2$

$=4a^2({\alpha }^2-{\beta }^2-2{\alpha }^2)+4a^2({\alpha }^2+{\beta }^2)+({\alpha }^2+{\beta }^2{)}^2$

$=4a^2(a^2-2{\alpha }^2)+4a^2({\alpha }^2+{\beta }^2)+(a^2+{\beta }^2{)}^2$

$=4a^2({\alpha }^2+{\beta }^2-2{\alpha }^2)+4a^2({\alpha }^2+{\beta }^2)+({\alpha }^2+{\beta }^2{)}^2$

$=-4a^2(a^2+b^2)+4a^2({\alpha }^2+{\beta }^2)+({\alpha }^2+{\beta }^2{)}^2$

$=({\alpha }^2+{\beta }^2{)}^2$

$=C{P}^4$

$\therefore SP.S^{\prime }P=C{P}^2$