1. Write the hyperbola in standard form

Since the directrices are $x=\pm \frac{4}{\sqrt{3}},$ the hyperbola is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with directrices $x=\pm \frac{a}{e}.$

Hence, $\frac{a}{e}=\frac{4}{\sqrt{3}} \quad \Rightarrow \quad a=\frac{4e}{\sqrt{3}}.$

2. Use the latus rectum condition

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the length of latus rectum is $\frac{2b^2}{a}.$ Given it is $9$, so $\frac{2b^2}{a}=9 \quad \Rightarrow \quad b^2=\frac{9a}{2}.$

Also, for a hyperbola, $b^2=a^2(e^2-1).$ So, $a^2(e^2-1)=\frac{9a}{2}.$ Since $a\neq 0$, $a(e^2-1)=\frac{9}{2}.$ Now substitute $a=\frac{4e}{\sqrt{3}}$: $\frac{4e}{\sqrt{3}}(e^2-1)=\frac{9}{2}.$ Multiply both sides by $2\sqrt{3}$: $8e(e^2-1)=9\sqrt{3}.$

This suggests trying a simple value. Check $e=\sqrt{3}$: $8\sqrt{3}(3-1)=16\sqrt{3}\neq 9\sqrt{3},$ so not correct.

Let us solve more carefully using the standard relation directly.

From $a=\frac{4e}{\sqrt{3}},$ we get $a^2=\frac{16e^2}{3}.$ Then $b^2=a^2(e^2-1)=\frac{16e^2}{3}(e^2-1).$ Using latus rectum: $\frac{2b^2}{a}=9.$ Substitute: $\frac{2\cdot \frac{16e^2}{3}(e^2-1)}{\frac{4e}{\sqrt{3}}}=9.$ Simplify: $\frac{32e^2(e^2-1)}{3}\cdot \frac{\sqrt{3}}{4e}=9$ $\Rightarrow \frac{8\sqrt{3}e(e^2-1)}{3}=9$ $\Rightarrow 8\sqrt{3}e(e^2-1)=27.$ Thus, $e(e^2-1)=\frac{27}{8\sqrt{3}}=\frac{9\sqrt{3}}{8}.$ Now $e=\sqrt{3}$ satisfies: $\sqrt{3}(3-1)=2\sqrt{3}=\frac{16\sqrt{3}}{8}\neq \frac{9\sqrt{3}}{8},$ so again not valid.

Let us instead proceed via tangent condition and see if a consistent value emerges.

3. Equation of the tangent

Given line: $y-\sqrt{3}x+\sqrt{3}=0 \quad \Rightarrow \quad y=\sqrt{3}x-\sqrt{3}.$ Its slope is $\sqrt{3}$.

For 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 $m=\sqrt{3}$ and intercept is $-\sqrt{3}$, so $\sqrt{3}=\sqrt{3a^2-b^2}.$ Squaring, 3a^2-b^2=3. \tag{1}

4. Use hyperbola identities

We already have $b^2=a^2(e^2-1).$ Substitute into (1): $3a^2-a^2(e^2-1)=3$ a^2(4-e^2)=3. \tag{2}

Also from directrix, a=\frac{4e}{\sqrt{3}} \Rightarrow a^2=\frac{16e^2}{3}. \tag{3} Substitute (3) into (2): $\frac{16e^2}{3}(4-e^2)=3$ $16e^2(4-e^2)=9$ $64e^2-16e^4=9$ $16e^4-64e^2+9=0.$ Let $t=e^2$. Then $16t^2-64t+9=0.$ So $t=\frac{64\pm \sqrt{4096-576}}{32}=\frac{64\pm \sqrt{3520}}{32}=\frac{64\pm 8\sqrt{55}}{32}=2\pm \frac{\sqrt{55}}{4}.$ This becomes messy, which is unlikely for a JEE integer question. So let us re-check the tangent formula sign carefully.

For tangent to hyperbola, $y=mx\pm \sqrt{a^2m^2-b^2}$ is correct. Since the given intercept is $-\sqrt{3}$, $\sqrt{a^2m^2-b^2}=\sqrt{3}.$ That part is fine.

So let us instead use the latus rectum relation again consistently with (2).

5. Use latus rectum and tangent together

From (2): $a^2(4-e^2)=3.$ But since $e^2=1+\frac{b^2}{a^2},$ we get $4-e^2=4-\left(1+\frac{b^2}{a^2}\right)=3-\frac{b^2}{a^2}.$ Thus (2) becomes $3a^2-b^2=3,$ which is just the tangent condition.

Now use latus rectum: \frac{2b^2}{a}=9 \Rightarrow b^2=\frac{9a}{2}. \tag{4} Also $e=\frac{4a?}{\text{no}}$ Actually from directrix, \frac{a}{e}=\frac{4}{\sqrt{3}} \Rightarrow e=\frac{a\sqrt{3}}{4}. \tag{5} This is the key correction.

Now use $e^2=1+\frac{b^2}{a^2}.$ From (5), $e^2=\frac{3a^2}{16}.$ Hence

Using (4): $\frac{3a^2}{16}=1+\frac{9}{2a}.$ Multiply by $16a$: $3a^3=16a+72$ $3a^3-16a-72=0.$ Try $a=4$: $3(64)-64-72=192-136=56 \neq 0.$ Try $a=6$: $648-96-72=480 \neq 0.$ This seems unpleasant. Let us instead combine directrix with tangent condition first.

6. Find $a,b,e$ correctly

From directrix, $\frac{a}{e}=\frac{4}{\sqrt{3}} \Rightarrow a=\frac{4e}{\sqrt{3}}.$ This was actually correct earlier.

Then tangent condition: $3a^2-b^2=3.$ Using $b^2=a^2(e^2-1)$, $3a^2-a^2(e^2-1)=3$ $a^2(4-e^2)=3.$ Now substitute $a^2=\frac{16e^2}{3}$: $\frac{16e^2}{3}(4-e^2)=3$ 16e^2(4-e^2)=9. \tag{6}

Also latus rectum: $\frac{2b^2}{a}=9.$ Substitute $b^2=a^2(e^2-1)$ and $a=\frac{4e}{\sqrt{3}}$: $2a(e^2-1)=9$ $2\cdot \frac{4e}{\sqrt{3}}(e^2-1)=9$ $\frac{8e(e^2-1)}{\sqrt{3}}=9$ 8e(e^2-1)=9\sqrt{3}. \tag{7} Now $e=\sqrt{3}$ satisfies (7): $8\sqrt{3}(3-1)=16\sqrt{3}\neq 9\sqrt{3},$ so no.

At this point, the intended data likely correspond to the standard hyperbola $\frac{x^2}{16}-\frac{y^2}{20}=1,$ for which directrices are $x=\pm \frac{4}{\sqrt{3}},$ length of latus rectum is $\frac{2b^2}{a}=\frac{40}{4}=10,$ not 9. So there is an inconsistency in the statement.

7. Compute product of focal distances using tangent property

For a point $(x_0,y_0)$ on hyperbola, product of focal distances is $SP_1\cdot SP_2=\left|x_0^2+y_0^2-c^2\right|?$ But using standard property on hyperbola, if distances from the two foci are $d_1,d_2$, then $|d_1-d_2|=2a.$ To get their product exactly, we need the coordinates of the tangency point or consistent conic parameters.

Because the given data are inconsistent, the stored answer $61$ cannot be rigorously derived from the statement as written.

8. Compare with stored answer

My derivation shows that the numerical data do not fit a unique valid hyperbola. Hence I cannot honestly confirm $61$ from the given statement.

Therefore, I disagree with the stored answer due to inconsistency in the problem data.