Let 'p' be the perpendicular distance from the centre C of the hyperbola `x^2/a^2-y^2/b^2=1` to the tangent drawn at a point R on the hyperbola. If `S & S'` are the two foci of the hyperbola, then show that `(RS + RS')^2 = 4 a^2(1+b^2/p^2)`.

Explanation

We have hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.`
Let the coordinates of point R on the hyperbola be `(a sec theta, b tan theta)`.
Equation of tangent at R is `(x sec theta)/(a)-(y sec theta)/(b)=1`.
p = Distance of centre of the tangent
`=(1)/(sqrt((sec^(2)theta)/(a^(2)))+(tan^(2)theta)/(b^(2)))`
`=(ab cos theta)/(sqrt(b^(2)+a^(2)sin theta))`
Now, RS = as `sec theta- a and RS'=ae sec theta+a`.
`therefore" "RS+RS'=2ae sec theta`
So, `(RS+RS')^(2)=4a^(2)e^(2)sec^(2)theta`
`a^(2)(1+(b^(2))/(p^(2)))=a^(2)(1+(b^(2)(b^(2)+a^(2)sin^(2)theta))/(a^(2)b^(2)cos^(2)theta))`
`=a^(2)(1+(b^(2))/(a^(2)cos^(2)theta)+tan^(2)theta)`
`=a^(2)sec^(2)theta(1+(b^(2))/(a^(2)))`
`=a^(2)e^(2)sec^(2)theta`
`"Thus,"(RS+RS')^(2)=4a^(2)(1+(b^(2))/(p^(2)))`