The tangent at an extremity (in the first quadrant) of the latus...

Hyperbola, $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1 ⇒ \dfrac{{x}^{2}}{4} - \dfrac{{y}^{2}}{5} = 1$

$\text{∴}{a}^{2} = 4, {b}^{2} = 5$

${b}^{2} = {a}^{2}\left({e}^{2} - 1\right) ⇒ 5 = 4\left({e}^{2} - 1\right)$

$⇒ {e}^{2} = \dfrac{5}{4} + 1 = \dfrac{9}{4}$

$⇒ e = \dfrac{3}{2}$

$∴$ Extremity of $LR$ in first quadrant $\text{≡}L\text{≡}\left(ae, \dfrac{{b}^{2}}{a}\right)$

$≡ \left(3\text{,}\dfrac{5}{2}\right)$

Equation of tangent to the hyperbola $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$ at \left({x}_{1}, {y}_{1}\right)$is$\dfrac{x x{}_{1}}{{a}^{2}} - \dfrac{y y{}_{1}}{{b}^{2}} = 1$</p><p>$⇒ \dfrac{x \cdot 3}{4} - \dfrac{y \cdot \left(\dfrac{5}{2}\right)}{5} = 1$</p><p>$⇒ \dfrac{3x }{4} - \dfrac{y }{2} = 1$</p><p>To find the point$A$on$x$--axis, put$y = 0$</p><p>$⇒ x = \dfrac{4}{3}$</p><p>$⇒ OA = \dfrac{4}{3}$</p><p>To find the point$B$on$y$-axis, put$x = 0$</p><p>$⇒ y = - 2$</p><p>$⇒ OB = 2$</p><p>Now,$O{A}^{2} - O{B}^{2} = \dfrac{16}{9} - 4 = - \dfrac{20}{9} .