DTP REQUIRED...

$\mathrm{eccentricity} \text{of} \mathrm{hyperbola}=\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

Distance $AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\Rightarrow \sqrt{{\left(ae-ae\right)}^2+{\left(\frac{-b^2}{a}-\frac{b^2}{a}\right)}^2}$

$\Rightarrow \sqrt{\frac{4b^4}{a^2}}=\frac{2b^2}{a}$

Now, $\tan 45°=\frac{b^2}{a}$

$1=\frac{{\displaystyle \frac{b^2}{a}}}{ae}$

$\Rightarrow e=\frac{b^2}{a^2}$

Now, $e=\sqrt{1+\frac{b^2}{a^2}}$

$e^2=1+\frac{b^2}{a^2}$

$\Rightarrow \frac{b^2}{a^2}=e^2-1$

$e=e^2-1$

$\Rightarrow e^2-e-1=0$

Which is quadratic equation compare it with $a{x}^2+bx+c$

We get $\mathrm{a}=1,\mathrm{b}=-1,\mathrm{c}=-1$

$e=\frac{-1\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-\left(-1\right)\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(-1\right)}}{2\times 1}$

$e=\frac{1\pm \sqrt{5}}{2}$

Eccentricity is always positive

$\therefore e=\frac{1+\sqrt{5}}{2}$