Explanation

(a)
Given, $( \pm ae,\ 0) = ( \pm 3,\ 0)$
$\Rightarrow \ \ ae = 3$
$\Rightarrow \ a^{2}e^{2} = 9\$
${{\Rightarrow \ }b}^{2} + a^{2} = 9$ …(i)
$\because\ 2x + y - 4 = 0$
$\Rightarrow \ \ \ y = - 2x + 4$
is the tangent to the hyperbola
$\therefore\ \ (4)^{2} = a^{2}( - 2)^{2} - b^{2}$
$\Rightarrow \ \ 4a^{2} - b^{2} = 16$ …(ii)
On solving Eqs. (i) and (ii), we get
$a^{2} = 5,\ b^{2} = 4\$
$\therefore$ Equation of hyperbola is $\frac{x^{2}}{5}–\frac{y^{2}}{4}\ = 1$
$\Rightarrow \ 4x^{2} - 5y^{2} = 20$