Question

If the normal at \theta on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the transverse axis at G, then AG.$A^{{}^{\prime }}$G is (where, A and $A^{{}^{\prime }}$ are the vertices of the hyperbola)

• A.
  
• B.
  
• C.
• D. None of the above

Answer 1

The correct option is C

The equation of the normal at $(asec\theta ,btan\theta )$ to the given hyperbola is ax $cos\theta +bycot\theta =(a^2+b^2)$ This meets the transverse axis i.e. X-axis at G. So, the coordinates of G are $(\frac{a^2+b^2}{a}sec\theta ,0)$
The coordinates of the vertices A and $A^{{}^{\prime }}$ are A(a, 0) and $A^{{}^{\prime }}(-a,\theta )$, respectively.
$\therefore$ $AG.A^{{}^{\prime }}G=(-a+\frac{a^2+b^2}{a}sec\theta )(a+\frac{a^2+b^2}{a}sec\theta ,)$
$=(-a+a{e}^{2}sec\theta )(a+a{e}^{2}sec\theta )=a^2(e^{4}sec2\theta -1)$