If the normal at `'theta'` on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` meets the transverse axis at `G`, and `A` and `A'` are the vertices of the hyperbola , then `AC.A'G=` (a) `a^2(e^4 sec^2 theta-1)` (b) `a^2(e^4 tan^2 theta-1)` (c) `b^2(e^4 sec^2 theta-1)` (d) `b^2(e^4 sec^2 theta+1)`

Options

A. `a^(2)(e^(2)sec^(2)theta-1)`
B. `a^(2)(e^(4)sec^(2)theta-1)`
C. `a^(2)(e^(4)sec^(2)theta+1)`
D. none of these

Explanation

The equation of the normal at `(asec theta, b tan theta)` to the given hyperbola is
`axcostheta+by cot theta=(a^(2)+b^(2))`
This meets the transverse axis i.e.`x`-axis at `G`. So, the coordinates of `G` are `((a^(2)+b^(2))/(a)sectheta,0)`.
The coordinates of the vertices `A` and `A'` are `A(a,0)` and `A'(-a,0)` respectively.
`:.AG.A'G=(-a+(a^(2)+b^(2))/(a)sectheta)(a+(a^(2)+b^(2))/(a)sectheta)`
`implies AG.A'G=(-a+ae^(2)sectheta)(a+ae^(2)sectheta)=a^(2)(e^(4)sec^(2)theta-1)`