Question

Sol. [Key Concept: Substitute the given point and solve $15e^{2} - 34e + 15 = 0$ for $e = \frac{5}{3}$; get $b^{2}/a^{2}$, then the latus rectum $\frac{2 \cdot 2(a^{2} + 1)}{b}$ of the second hyperbola.]

$$\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 $$It passs through $(6,4\sqrt{3})$

$$\begin{array}{r} \Rightarrow \frac{36}{a^{2}} - \frac{48}{{\text{ }b}^{2}} = 1\#(1) \end{array}$$

also $15e^{2} - 34e + 15 = 0$

$$\Rightarrow 15e^{2} - 25e - 9e + 15 = 0 $$$e = \frac{5}{3}$ or $\frac{3}{5} \Rightarrow e = \frac{5}{3}$

$$\begin{array}{r} 1 + \frac{b^{2}}{a^{2}} = \frac{25}{9} \Rightarrow \frac{{\text{ }b}^{2}}{a^{2}} = \frac{16}{9}\#(2) \end{array}$$

using (1) and (2)

$$\Rightarrow \frac{36}{a^{2}} - \frac{48}{16a^{2}} \times 9 = 1 \Rightarrow a = 3,b = 4 $$Length of L.R.

$$\frac{4\left( a^{2} + 1 \right)}{b} = \frac{4(10)}{4} = 10 $$