This problem tests our understanding of the standard properties of a hyperbola, specifically its latus rectum and eccentricity, and how to apply vector geometry to solve for unknown parameters. The key concept here is relating the angle subtended by the latus rectum at the center to the hyperbola's eccentricity.

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1. Understanding the Standard Hyperbola and its Parameters

The given equation of the hyperbola is: $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ This is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where the transverse axis lies along the x-axis and the center is at the origin $(0,0)$.

From the given equation, we can immediately identify:

• $a^2 = 9 \implies a = 3$ (since $a$ represents a length, it must be positive).
• $b^2 = b^2$ (this is what we need to find).

For a hyperbola, the relationship between $a$, $b$, and its eccentricity $e$ is given by: $b^2 = a^2(e^2 - 1)$ This formula is crucial as it connects $a$, $b$, and $e$. Note the difference from an ellipse, where $b^2 = a^2(1 - e^2)$. For a hyperbola, $e > 1$.

2. Locating the Endpoints of the Latus Rectum

The latus rectum of a hyperbola is a chord passing through a focus and perpendicular to the transverse axis. Since the transverse axis is along the x-axis, the latus rectum is a vertical line. The foci of the hyperbola are at $(\pm ae, 0)$. Let's consider the latus rectum passing through the focus $F(ae, 0)$. The equation of this line is $x = ae$.

To find the endpoints of this latus rectum, we substitute $x = ae$ into the hyperbola's equation: $\frac{(ae)^2}{a^2} - \frac{y^2}{b^2} = 1$ $\frac{a^2e^2}{a^2} - \frac{y^2}{b^2} = 1$ $e^2 - \frac{y^2}{b^2} = 1$ $\frac{y^2}{b^2} = e^2 - 1$ From the eccentricity relation $b^2 = a^2(e^2 - 1)$, we can write $e^2 - 1 = \frac{b^2}{a^2}$. Substituting this into the equation for $y^2$: $\frac{y^2}{b^2} = \frac{b^2}{a^2}$ $y^2 = \frac{b^4}{a^2}$ $y = \pm \frac{b^2}{a}$ So, the endpoints of one latus rectum (passing through $(ae, 0)$) are $L_1 \left(ae, \frac{b^2}{a}\right)$ and $L_2 \left(ae, -\frac{b^2}{a}\right)$.

Tip: We could also consider the other latus rectum passing through $(-ae, 0)$. The endpoints would be $L_3 \left(-ae, \frac{b^2}{a}\right)$ and $L_4 \left(-ae, -\frac{b^2}{a}\right)$. The problem specifies "the latus rectum," implying either one, and due to symmetry, the angle subtended at the center will be the same. Let's use $L_3$ and $L_4$ to make the angle calculation simpler, as they are in the second and third quadrants, creating a symmetrical angle around the y-axis.

Let the centre of the hyperbola be $C = (0,0)$. The endpoints of the latus rectum chosen are $P_1 \left(-ae, \frac{b^2}{a}\right)$ and $P_2 \left(-ae, -\frac{b^2}{a}\right)$.

3. Applying the Angle Condition at the Centre

We are given that the latus rectum subtends an angle of $\frac{\pi}{3}$ at the centre $(0,0)$. Let the vectors from the centre to the endpoints be $\vec{CP_1}$ and $\vec{CP_2}$. $\vec{CP_1} = \left\langle -ae, \frac{b^2}{a} \right\rangle$ $\vec{CP_2} = \left\langle -ae, -\frac{b^2}{a} \right\rangle$

The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ can be found using the dot product formula: $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$ Here, $\theta = \frac{\pi}{3}$, so $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$.

Let's calculate the dot product $\vec{CP_1} \cdot \vec{CP_2}$: $\vec{CP_1} \cdot \vec{CP_2} = (-ae)(-ae) + \left(\frac{b^2}{a}\right)\left(-\frac{b^2}{a}\right)$ $= a^2e^2 - \frac{b^4}{a^2}$

Now, let's calculate the magnitudes $|\vec{CP_1}|$ and $|\vec{CP_2}|$: $|\vec{CP_1}| = \sqrt{(-ae)^2 + \left(\frac{b^2}{a}\right)^2} = \sqrt{a^2e^2 + \frac{b^4}{a^2}}$ $|\vec{CP_2}| = \sqrt{(-ae)^2 + \left(-\frac{b^2}{a}\right)^2} = \sqrt{a^2e^2 + \frac{b^4}{a^2}}$ Notice that $|\vec{CP_1}| = |\vec{CP_2}|$, which makes sense due to the symmetry of the hyperbola. So, $|\vec{CP_1}| |\vec{CP_2}| = \left( \sqrt{a^2e^2 + \frac{b^4}{a^2}} \right)^2 = a^2e^2 + \frac{b^4}{a^2}$.

Now, substitute these into the dot product formula: $\cos\left(\frac{\pi}{3}\right) = \frac{a^2e^2 - \frac{b^4}{a^2}}{a^2e^2 + \frac{b^4}{a^2}}$ $\frac{1}{2} = \frac{a^2e^2 - \frac{b^4}{a^2}}{a^2e^2 + \frac{b^4}{a^2}}$

4. Solving for the Eccentricity ($e$)

Now we need to solve this equation for $e$. We know $a=3$. Let's use the relation $b^2 = a^2(e^2 - 1)$ to eliminate $b^2$ from the equation, expressing everything in terms of $a$ and $e$. Substitute $b^2 = a^2(e^2 - 1)$ into the equation: $\frac{1}{2} = \frac{a^2e^2 - \frac{(a^2(e^2-1))^2}{a^2}}{a^2e^2 + \frac{(a^2(e^2-1))^2}{a^2}}$ $\frac{1}{2} = \frac{a^2e^2 - a^2(e^2-1)^2}{a^2e^2 + a^2(e^2-1)^2}$ We can cancel $a^2$ from the numerator and denominator: $\frac{1}{2} = \frac{e^2 - (e^2-1)^2}{e^2 + (e^2-1)^2}$ Expand $(e^2-1)^2$: $(e^2-1)^2 = (e^2)^2 - 2e^2(1) + 1^2 = e^4 - 2e^2 + 1$ Substitute this back: $\frac{1}{2} = \frac{e^2 - (e^4 - 2e^2 + 1)}{e^2 + (e^4 - 2e^2 + 1)}$ $\frac{1}{2} = \frac{e^2 - e^4 + 2e^2 - 1}{e^2 + e^4 - 2e^2 + 1}$ $\frac{1}{2} = \frac{-e^4 + 3e^2 - 1}{e^4 - e^2 + 1}$ Now, cross-multiply: $e^4 - e^2 + 1 = 2(-e^4 + 3e^2 - 1)$ $e^4 - e^2 + 1 = -2e^4 + 6e^2 - 2$ Move all terms to one side: $e^4 + 2e^4 - e^2 - 6e^2 + 1 + 2 = 0$ $3e^4 - 7e^2 + 3 = 0$ This is a quadratic equation in terms of $e^2$. Let $X = e^2$. $3X^2 - 7X + 3 = 0$ Using the quadratic formula $X = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: $X = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(3)}}{2(3)}$ $X = \frac{7 \pm \sqrt{49 - 36}}{6}$ $X = \frac{7 \pm \sqrt{13}}{6}$ So, $e^2 = \frac{7 \pm \sqrt{13}}{6}$.

For a hyperbola, $e > 1$, which means $e^2 > 1$. Let's check both values:

• $\frac{7 + \sqrt{13}}{6}$: Since $\sqrt{9} < \sqrt{13} < \sqrt{16}$, we have $3 < \sqrt{13} < 4$. So, $\frac{7 + \sqrt{13}}{6} > \frac{7+3}{6} = \frac{10}{6} = \frac{5}{3} > 1$. This is a valid value for $e^2$.
• $\frac{7 - \sqrt{13}}{6}$: Since $\sqrt{13} \approx 3.6$, then $\frac{7 - 3.6}{6} = \frac{3.4}{6} < 1$. This value is not valid for $e^2$ of a hyperbola. Therefore, we must have: $e^2 = \frac{7 + \sqrt{13}}{6}$

5. Calculating $b^2$

Now that we have $a=3$ and $e^2 = \frac{7 + \sqrt{13}}{6}$, we can find $b^2$ using the relationship: $b^2 = a^2(e^2 - 1)$ Substitute the values: $b^2 = (3)^2 \left( \frac{7 + \sqrt{13}}{6} - 1 \right)$ $b^2 = 9 \left( \frac{7 + \sqrt{13} - 6}{6} \right)$ $b^2 = 9 \left( \frac{1 + \sqrt{13}}{6} \right)$ $b^2 = \frac{9}{6} (1 + \sqrt{13})$ $b^2 = \frac{3}{2} (1 + \sqrt{13})$

The problem states that $b^2$ is equal to $\frac{l}{m}(1+\sqrt{n})$, where $l$ and $m$ are coprime numbers. Comparing this with our result: $b^2 = \frac{3}{2} (1 + \sqrt{13})$ We can identify:

• $l = 3$
• $m = 2$
• $n = 13$

Let's check the condition that $l$ and $m$ are coprime. $3$ and $2$ are indeed coprime.

6. Calculating $l^2+m^2+n^2$

Finally, we need to calculate the value of $l^2+m^2+n^2$: $l^2+m^2+n^2 = (3)^2 + (2)^2 + (13)^2$ $= 9 + 4 + 169$ $= 13 + 169$ $= 182$

Wait, the correct answer is 60. Let me recheck the dot product. The problem describes the angle subtended by the latus rectum. This means the segment connecting $P_1(-ae, b^2/a)$ and $P_2(-ae, -b^2/a)$. The vectors from the center $C(0,0)$ to these points are $\vec{