The hyperbola $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ passes through the point of intersection of the lines $x-3\sqrt { 5 } y=0$ and $\sqrt { 5 } x-2y=13$ and the length of its flatus rectum is 4/3 units. The coordinates of its focus are-

To find the coordinates of the focus of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we need to determine $a, b$, and the coordinates of the point through which the hyperbola passes.

First, find the point of intersection of the lines $x-3 \sqrt{5} y=0$ and $\sqrt{5} x-$ $2 y=13$.

1. Solve the first equation for $x$ : $x=3 \sqrt{5} y$

2. Substitute $x=3 \sqrt{5} y$ into the second equation:

$\begin{array}{l} \sqrt{5}(3 \sqrt{5} y)-2 y=13 \\ 15 y-2 y=13 \\ 13 y=13 \\ y=1 \end{array}$

3. Substitute $y=1$ back into $x=3 \sqrt{5} y$ :

$\begin{array}{l} x=3 \sqrt{5} \cdot 1 \\ x=3 \sqrt{5} \end{array}$

Thus, the point of intersection is $(3 \sqrt{5}, 1)$.

Next, we use the given condition that the length of the latus rectum of the hyperbola is $\frac{4}{3}$. For a hyperbola, the length of the latus rectum is given by: $\frac{2 b^{2}}{a}=\frac{4}{3}$

Thus:

$\begin{array}{l} 2 b^{2}=\frac{4 a}{3} \\ 6 b^{2}=4 a \\ b^{2}=\frac{2 a}{3} \end{array}$

Now,

we know that the point $(3 \sqrt{5}, 1)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

Substitute $x=3 \sqrt{5}$ and $y=1$ into this equation:

$\begin{array}{l} \frac{(3 \sqrt{5})^{2}}{a^{2}}-\frac{1^{2}}{b^{2}}=1 \\ \frac{45}{a^{2}}-\frac{1}{b^{2}}=1 \end{array}$

Substitute $b^{2}=\frac{2 a}{3}$ into the equation:

$\begin{array}{l} \frac{45}{a^{2}}-\frac{1}{\frac{2 a}{3}}=1 \\ \frac{45}{a^{2}}-\frac{3}{2 a}=1 \\ \frac{45}{a^{2}}-\frac{3}{2 a}=1 \end{array}$

Multiply through by $2 a^{2}$ to clear the denominators:

$\begin{array}{l} 90-3 a=2 a^{2} \\ 2 a^{2}+3 a-90=0 \end{array}$

Solve this quadratic equation for $a$ :

$a=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

where $a=2, b=3$, and $c=-90$ :

$\begin{array}{l} a=\frac{-3 \pm \sqrt{9+720}}{4} \\ a=\frac{-3 \pm \sqrt{729}}{4} \\ a=\frac{-3 \pm 27}{4} \end{array}$

So,

$a=\frac{24}{4}=6$

or,

$a=\frac{-30}{4}=-7.5$

Since $a$ represents a distance and must be positive:

$a=6$

Now, find $b$ :

$\begin{array}{l} b^{2}=\frac{2 a}{3}=\frac{2 \cdot 6}{3}=4 \\ b=2 \end{array}$

Finally, the coordinates of the foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are given by $( \pm c, 0)$, where $c=\sqrt{a^{2}+b^{2}}$ :

$c=\sqrt{6^{2}+2^{2}}=\sqrt{36+4}=\sqrt{40}=2 \sqrt{10}$

Therefore, the coordinates of the foci are: $( \pm 2 \sqrt{10}, 0)$