1. Given hyperbola

$\frac{x^2}{9}-\frac{y^2}{4}=1$

This is of the form

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

with

$a^2=9,\quad b^2=4.$

2. Slope of the given line

The line is

$2x-y=1 \implies y=2x-1$

So its slope is

$m=2.$

Since the tangents are parallel to this line, the tangents to the hyperbola must also have slope $2$.

3. Equation of tangent to hyperbola with slope $m$

For the hyperbola

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

a tangent with slope $m$ is

$y=mx\pm \sqrt{a^2m^2-b^2}.$

Here $a^2=9$, $b^2=4$, $m=2$. Hence

$y=2x\pm \sqrt{9\cdot 4-4}=2x\pm \sqrt{36-4}=2x\pm \sqrt{32}=2x\pm 4\sqrt 2.$

So the two tangents are

$y=2x+4\sqrt2$

and

$y=2x-4\sqrt2.$

4. Find point of contact corresponding to slope $2$

Differentiate the hyperbola implicitly:

$\frac{x^2}{9}-\frac{y^2}{4}=1$

$\Rightarrow \frac{2x}{9}-\frac{2y}{4}\frac{dy}{dx}=0$

$\Rightarrow \frac{2x}{9}-\frac{y}{2}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}.$

For tangent slope $2$,

$\frac{4x}{9y}=2$

$\Rightarrow 4x=18y$

$\Rightarrow 2x=9y$

$\Rightarrow y=\frac{2x}{9}.$

5. Substitute into the hyperbola

Put $y=\dfrac{2x}{9}$ into

$\frac{x^2}{9}-\frac{y^2}{4}=1.$

Then

$\frac{x^2}{9}-\frac{1}{4}\left(\frac{2x}{9}\right)^2=1$

$\frac{x^2}{9}-\frac{1}{4}\cdot \frac{4x^2}{81}=1$

$\frac{x^2}{9}-\frac{x^2}{81}=1$

$\frac{9x^2-x^2}{81}=1$

$\frac{8x^2}{81}=1$

$x^2=\frac{81}{8}$

$x=\pm \frac{9}{2\sqrt2}.$

Now

$y=\frac{2x}{9}=\pm \frac{1}{\sqrt2}.$

So the points of contact are

$\left(\frac{9}{2\sqrt2},\frac{1}{\sqrt2}\right)$

and

$\left(-\frac{9}{2\sqrt2},-\frac{1}{\sqrt2}\right).$

6. Match with options

• A: $\left( \frac{9}{2\sqrt 2 },\frac{1}{\sqrt 2 } \right)$ ✓
• B: $\left( -\frac{9}{2\sqrt 2 },-\frac{1}{\sqrt 2 } \right)$ ✓
• C: $\left( 3\sqrt 3 , - 2\sqrt 2 \right)$ ✗
• D: $\left( -3\sqrt 3 , 2\sqrt 2 \right)$ ✗

Therefore, the correct options are A and B.