Let m _ 1 and m _ 2 be the slopes of the tangents drawn from the...

Given that $H: \dfrac{{y}^{2}}{25} - \dfrac{{x}^{2}}{16} = 1$

Equation of tangent to the hyperbola is$: y = mx + \sqrt{25 - 16{m}^{2}}$

Passes through \left(4, 1\right)$</p><p>$⇒ y = mx + \sqrt{25 - 16{m}^{2}}$</p><p>$⇒ 1 = 4m + \sqrt{25 - 16{m}^{2}}$</p><p>$⇒ 4{m}^{2} - m - 3$</p><p>$⇒ {m}_{1} = 1, {m}_{2} = \dfrac{- 3}{4}$</p><p>$⇒ \left|{m}_{1}\right| = 1, \left|{m}_{2}\right| = \dfrac{3}{4}$</p><p>Equation of tangents with slopes$1&\dfrac{3}{4}$</p><p>$⇒ y = x - 3 . . . . . . . \left(i\right)$</p><p>$⇒ y = \dfrac{3}{4}x - 4 . . . . . . \left(ii\right)$</p><p>\left(i\right)$ & \left(ii\right)$⇒ Q ≡ ( - 4, - 7 )$

$P ≡ ( 4, 1 )$

$⇒ ( PQ{)}^{2} = {8}^{2} + {8}^{2} = 128$

On converting \left(i\right)$and \left(ii\right)$ in intercept form we get,

\left(i\right) ⇒ \alpha = 3, \left(ii\right) ⇒ \beta = \dfrac{16}{3}$</p><p>$⇒ \dfrac{( PQ{)}^{2}}{\alpha\beta} = \dfrac{128}{3 \times \dfrac{16}{3}} = 8

Hence this is the required option.