The foci of a hyperbola are ( ± 2, 0 ) and its eccentricity is...

The given equation of hyperbola is $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1, ae = 2, e = \dfrac{3}{2}$

$⇒ a = \dfrac{4}{3}$

$⇒ {b}^{2} = {a}^{2}{e}^{2} - {a}^{2} = 4 - \dfrac{16}{9} = \dfrac{20}{9}$

The tangent is perpendicular to $2x + 3y = 6$.

Slope of tangent is $m = \dfrac{- 1}{\dfrac{- 2}{3}} = \dfrac{3}{2}$

Equation of tangent, $m = \dfrac{3}{2}$

Equation of tangent is $y = mx - \sqrt{{a}^{2}{m}^{2} - {b}^{2}}$
$⇒ y = \dfrac{3}{2}x - \sqrt{\dfrac{16}{9} \times \dfrac{9}{4} - \dfrac{20}{9}}$

( $∵$ Tangent is in ${1}^{\text{st}}$ quadrant\left⇒ C < 0\right)$</p><p>$⇒ y = \dfrac{3}{2}x - \dfrac{4}{3}$</p><p>Converting the equation into intercept form:</p><p>$⇒ \dfrac{x}{\left(\dfrac{8}{9}\right)} + \dfrac{y}{\left(- \dfrac{4}{3}\right)} = 1$</p><p>$⇒ \left|6a\right| + \left|5b\right| = \left|\dfrac{6 \times 8}{9}\right| + \left|\dfrac{5 \times \left(- 4\right)}{3}\right| = 12

Hence this is the required option.