Given hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

For this standard hyperbola:

• Foci are at $(\pm ae,0)$
• Directrices are $x=\pm \frac{a}{e}$
• Length of latus rectum is $l=\frac{2b^2}{a}$

Use the given focus

One focus is at $(\sqrt{10},0)$, so $ae=\sqrt{10} \qquad (1)$

Use the given corresponding directrix

The corresponding directrix is $x=\frac{9}{\sqrt{10}}$ so $\frac{a}{e}=\frac{9}{\sqrt{10}} \qquad (2)$

Find $a$ and $e$

Multiply (1) and (2): $\left(ae\right)\left(\frac{a}{e}\right)=\sqrt{10}\cdot \frac{9}{\sqrt{10}}$ $a^2=9$ $a=3$

From (1): $3e=\sqrt{10}$ $e=\frac{\sqrt{10}}{3}$ Hence, $e^2=\frac{10}{9}$

Find $b^2$

For hyperbola, $b^2=a^2(e^2-1)$ Therefore, $b^2=9\left(\frac{10}{9}-1\right)=9\cdot \frac{1}{9}=1$

Find latus rectum length

$l=\frac{2b^2}{a}=\frac{2\cdot 1}{3}=\frac{2}{3}$

Compute $9(e^2+l)$

$e^2+l=\frac{10}{9}+\frac{2}{3}$ $=\frac{10}{9}+\frac{6}{9}=\frac{16}{9}$

Thus, $9(e^2+l)=9\cdot \frac{16}{9}=16$

Check options

• A: 12
• B: 14
• C: 15
• D: 16

So the correct option is D.