This solution will guide you through the problem step-by-step, focusing on clarity, fundamental concepts, and common strategies useful for JEE Mathematics.

---

1. Understanding the Problem and Key Concepts

The problem asks for the square of the distance of a point $P$ from the origin. Point $P$ lies on a given hyperbola in the first quadrant, and we are provided with the area of the triangle formed by $P$ and the two foci of the hyperbola.

To solve this, we'll need to use:

• The standard equation and properties of a hyperbola to find its parameters ($a, b, e$) and the coordinates of its foci.
• The formula for the area of a triangle to determine a coordinate of point $P$.
• The fact that point $P$ lies on the hyperbola, meaning its coordinates must satisfy the hyperbola's equation.
• The distance formula (specifically, the square of the distance from the origin).

Let's begin by extracting information from the hyperbola's equation.

---

2. Analyzing the Hyperbola and Determining its Foci

The given hyperbola is $H: \frac{x^2}{9}-\frac{y^2}{4}=1$.

• Key Concept: The standard equation of a hyperbola centered at the origin, opening along the x-axis, is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

  • $a$ is the length of the semi-transverse axis.
  • $b$ is the length of the semi-conjugate axis.
  • The eccentricity $e$ is related by $b^2 = a^2(e^2-1)$ or $e^2 = 1 + \frac{b^2}{a^2}$.
  • The foci are located at $(\pm ae, 0)$.

• Step 2.1: Identify $a$ and $b$ from the hyperbola's equation. Comparing $\frac{x^2}{9}-\frac{y^2}{4}=1$ with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $a^2 = 9 \implies a = 3 \quad (\text{since } a > 0)$ $b^2 = 4 \implies b = 2 \quad (\text{since } b > 0)$

  • Why this step? The parameters $a$ and $b$ are fundamental to defining the hyperbola's shape and are essential for calculating its eccentricity and foci.

• Step 2.2: Calculate the eccentricity $e$. We use the relation $e^2 = 1 + \frac{b^2}{a^2}$: $e^2 = 1 + \frac{4}{9} = \frac{9+4}{9} = \frac{13}{9}$ Therefore, the eccentricity is: $e = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \quad (\text{since } e > 1 \text{ for a hyperbola})$

  • Why this step? Eccentricity is a crucial parameter for all conic sections. For a hyperbola, it's directly used to locate the foci.

• Step 2.3: Determine the coordinates of the foci. The foci are at $(\pm ae, 0)$. Let's calculate $ae$: $ae = 3 \times \frac{\sqrt{13}}{3} = \sqrt{13}$ So, the coordinates of the two foci are: $F_1 = (\sqrt{13}, 0) \quad \text{and} \quad F_2 = (-\sqrt{13}, 0)$

  • Why this step? These are the other two vertices of the triangle whose area is given. Knowing their coordinates is necessary to calculate the triangle's base.

---

3. Using the Area of the Triangle to Find a Coordinate of Point P

Let $P=(x_P, y_P)$ be the point on the hyperbola in the first quadrant. The triangle is formed by $P$, $F_1(\sqrt{13}, 0)$, and $F_2(-\sqrt{13}, 0)$. The area of this triangle is given as $2\sqrt{13}$.

• Key Concept: The area of a triangle can be calculated using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. This is particularly efficient when the base lies on one of the coordinate axes.

• Step 3.1: Identify the base and height of $\triangle PF_1F_2$. The foci $F_1$ and $F_2$ lie on the x-axis. Therefore, the segment $F_1F_2$ can be taken as the base of the triangle. $\text{Base} = F_1F_2 = \sqrt{13} - (-\sqrt{13}) = 2\sqrt{13}$ The height of the triangle with respect to this base is the perpendicular distance from point $P(x_P, y_P)$ to the x-axis, which is simply $|y_P|$. Since $P$ is in the first quadrant, $y_P > 0$, so the height is $y_P$.

  • Why this step? Choosing the base along the x-axis simplifies the height calculation significantly. This avoids more complex formulas like Heron's formula or determinant methods for area, making the solution more straightforward.

• Step 3.2: Set up and solve the area equation for $y_P$. We are given that the area of $\triangle PF_1F_2$ is $2\sqrt{13}$. Using the area formula: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$ $2\sqrt{13} = \frac{1}{2} \times (2\sqrt{13}) \times y_P$ $2\sqrt{13} = \sqrt{13} \times y_P$ Dividing both sides by $\sqrt{13}$ (which is non-zero): $y_P = 2$

  • Why this step? This calculation directly gives us the $y$-coordinate of point $P$, which is one of the two coordinates required to locate $P$.

---

4. Finding the $x$-coordinate of Point P

Now that we have $y_P=2$, we need to find $x_P$. Point $P(x_P, y_P)$ lies on the hyperbola, so its coordinates must satisfy the hyperbola's equation.

• Key Concept: Any point $(x,y)$ lying on a curve satisfies the equation of that curve.

• Step 4.1: Substitute $y_P=2$ into the hyperbola's equation. The equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$. Substitute $y_P=2$: $\frac{x_P^2}{9} - \frac{2^2}{4} = 1$ $\frac{x_P^2}{9} - \frac{4}{4} = 1$ $\frac{x_P^2}{9} - 1 = 1$

  • Why this step? By substituting the known $y_P$ value, we form an equation solely in terms of $x_P$, allowing us to solve for it.

• Step 4.2: Solve for $x_P^2$. Add 1 to both sides: $\frac{x_P^2}{9} = 1 + 1$ $\frac{x_P^2}{9} = 2$ Multiply both sides by 9: $x_P^2 = 18$ Since $P$ is in the first quadrant, $x_P > 0$, so $x_P = \sqrt{18} = 3\sqrt{2}$.

  • Why this step? We need $x_P^2$ for the final calculation of the square of the distance from the origin. The information about $P$ being in the first quadrant confirms that $x_P$ must be positive.

---

5. Calculating the Square of the Distance from the Origin

We have the coordinates of $P$ as $(x_P, y_P)$, where $x_P^2 = 18$ and $y_P = 2$. We need to find the square of the distance of $P$ from the origin $O(0,0)$.

• Key Concept: The square of the distance of a point $(x,y)$ from the origin $(0,0)$ is given by $x^2 + y^2$.

• Step 5.1: Use the distance formula. The square of the distance $OP^2$ is: $OP^2 = x_P^2 + y_P^2$ We found $x_P^2 = 18$ and $y_P = 2 \implies y_P^2 = 2^2 = 4$. $OP^2 = 18 + 4$ $OP^2 = 22$

  • Why this step? This is the final calculation that directly answers the question posed in the problem.

---

6. Important Tips and Common Pitfalls for JEE Aspirants

• Always Start with Parameters: For any conic section problem, the first crucial step is to correctly identify $a, b,$ and then calculate $e$ and the foci coordinates. This foundation prevents errors later.
• Geometric Visualization: Sketching the hyperbola, its foci, and the point $P$ can greatly aid in understanding the problem. In this case, realizing that the foci are on the x-axis immediately suggests using the base-height formula for the triangle's area, which is far simpler than other methods.
• Utilize Quadrant Information: The condition "in the first quadrant" is vital for determining the signs of $x_P$ and $y_P