1. Write the hyperbola in standard form

Given: $kx^2-y^2=6$

Divide by $6$: $\frac{kx^2}{6}-\frac{y^2}{6}=1$

Rewrite as $\frac{x^2}{6/k}-\frac{y^2}{6}=1$

So this is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a^2=\frac{6}{k}, \qquad b^2=6$

2. Use the directrix condition

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ the directrices are $x=\pm \frac{a}{e}$ where $e=\sqrt{1+\frac{b^2}{a^2}}$

Given directrix: $x-1=0 \Rightarrow x=1$

Hence, $\frac{a}{e}=1 \Rightarrow a=e$

Now, $e^2=1+\frac{b^2}{a^2}$ Since $a=e$, we get $a^2=1+\frac{b^2}{a^2}$

Using $b^2=6$: $a^2=1+\frac{6}{a^2}$

Multiply by $a^2$: $a^4-a^2-6=0$

Let $t=a^2$. Then $t^2-t-6=0$ $(t-3)(t+2)=0$

Since $a^2>0$, we take $a^2=3$

Thus, $\frac{6}{k}=3 \Rightarrow k=2$

3. Equation of hyperbola

Substitute $k=2$ into the original equation: $2x^2-y^2=6$

4. Check the options

We test each point in $2x^2-y^2=6$

• A: $(-2\sqrt{5},6)$ $2(20)-36=40-36=4 \ne 6$ Not on the hyperbola.

• B: $(-\sqrt{5},3)$ $2(5)-9=10-9=1 \ne 6$ Not on the hyperbola.

• C: $(\sqrt{5},-2)$ $2(5)-(-2)^2=10-4=6$ Lies on the hyperbola.

• D: $(2\sqrt{5},3\sqrt{6})$ $2(20)-(3\sqrt{6})^2=40-54=-14 \ne 6$ Not on the hyperbola.

5. Conclusion

The hyperbola passes through $\boxed{(\sqrt{5},-2)}$ which is Option C.