If two points `P & Q` on the hyperbola ,`x^2/a^2-y^2/b^2=1` whose centre is C be such that CP is perpendicularal to `CQ and a lt b`1 ,then prove that `1/(CP^2)+1/(CQ^2)=1/a^2-1/b^2`.

Options

A. `(b^(2)-a^(2))/(2ab)`
B. `(1)/(a^(2))+(1)/(b^(2))`
C. `(2ab)/(b^(2)-a^(2))`
D. `(1)/(a^(2))-(1)/(b^(2))`

Answer

D

Explanation

Let `CP=r_(1)` be inclined to the transverse axis at an angle `theta` so that P is (`r_(1) cos theta, r_(1) sin theta`) and P lies on the hyperbola. It gives
`r_(1)^(3)((cos^(2)theta)/(a^(2))-(sin^(2)theta)/(b^(2)))=1`

Replacing `theta` by `90^(@)+theta`, we have
`r_(2)^(2)((sin^(2)theta)/(a^(2))-(cos^(2)theta)/(b^(2)))=1`
`"or "(1)/(r_(1)^(3))+(1)/(r_(2)^(2))=(cos^(2)theta)/(a^(2))-(sin^(2)theta)/(b^(2))+(sin^(2)theta)/(a^(2))-(cos^(2))/(b^(2))`
`"or "(1)/(r_(1)^(2))+(1)/(r_(2)^(2))=cos^(2)theta((1)/(a^(2))-(1)/(b^(2)))+sin^(2)theta((1)/(a^(2))-(1)/(b^(2)))`
`"or "(1)/(r_(1)^(2))+(1)/(r_(2)^(2))=(1)/(a^(2))-(1)/(b^(2))`
`"or "(1)/(CP^(2))+(1)/(CQ^(2))=(1)/(a^(2))-(1)/(b^(2))`