Let P be a point on the hyperbola H : x²/9 - y²/4 = 1, in the first quadrant such that the area of t

Q: Wait, why is the base of the triangle $2ae$?

Great question! The foci of a standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are located at $(\pm ae, 0)$. The distance between $(ae, 0)$ and $(-ae, 0)$ is simply $ae - (-ae) = 2ae$. Since these points lie on the x-axis, they form a horizontal base for our triangle.

Q: Can I use the definition of a hyperbola (difference of focal distances) instead of the area formula?

While the definition of a hyperbola ($|PS_1 - PS_2| = 2a$) is powerful, it relates to the *lengths* of the sides, not the area of the triangle. The area formula is much more direct here because the base is fixed on the x-axis, making the height calculation trivial. Stick to the simplest path!

Q: Why do we only consider the positive value for $\beta$?

The problem explicitly states that point P is in the first quadrant. In the first quadrant, both the x-coordinate ($\alpha$) and the y-coordinate ($\beta$) must be positive. If we had found a negative value for $\beta$, we would have rejected it based on this constraint.

Question

Let P be a point on the hyperbola $H : \frac{x^2}{9} - \frac{y^2}{4} = 1$, in the first quadrant such that the area of triangle formed by P and the two foci of H is $2\sqrt{13}$. Then, the square of the distance of P from the origin is

Options:

(A) 18

(B) 26

(C) 22

(D) 20

Sigma Percentile · Accepted Answer

### The Hyperbola's Hidden Geometry

Welcome, fellow traveler, to the elegant world of conic sections. Today, we are not just solving an equation; we are uncovering the hidden geometry of a hyperbola.

Imagine standing on a coordinate plane, looking at the curve defined by:

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

This isn't just a set of points; it's a path, a trajectory that obeys strict mathematical laws. Our goal is to find a specific point $P$ on this curve that creates a triangle with the foci of the hyperbola, with an area of $2\sqrt{13}$.

### Decoding the Anatomy

Before we can find point $P$, we must understand the hyperbola itself. By comparing our equation to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we immediately see that $a^2 = 9$ and $b^2 = 4$.

This tells us that $a = 3$ and $b = 2$. These values are the DNA of our hyperbola, defining its shape and its reach.

Next, we need the eccentricity, $e$. The relationship for a hyperbola is $e^2 = 1 + \frac{b^2}{a^2}$. Substituting our values:

$$e^2 = 1 + \frac{4}{9} = \frac{13}{9} \Rightarrow e = \frac{\sqrt{13}}{3}$$

This value is crucial because it leads us to the foci, the two 'anchors' of the hyperbola. The foci are located at $(\pm ae, 0)$.

Calculating $ae$, we get $3 	imes \frac{\sqrt{13}}{3} = \sqrt{13}$. Thus, our foci are $S_1(-\sqrt{13}, 0)$ and $S_2(\sqrt{13}, 0)$.

### The Geometry of the Triangle

Now, imagine a point $P(\alpha, \beta)$ in the first quadrant. We connect this point to our two anchors, $S_1$ and $S_2$.

This forms a triangle, $	riangle PS_1S_2$, resting on the x-axis. The base of this triangle is the distance between the foci, which is $2ae = 2\sqrt{13}$.

The height of this triangle is simply the vertical distance from the x-axis to point $P$, which is the y-coordinate, $\beta$. The area of any triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.

Substituting our knowns:

$$	ext{Area} = \frac{1}{2} 	imes (2\sqrt{13}) 	imes \beta = \sqrt{13}\beta$$

We are given that this area is $2\sqrt{13}$. Equating the two, we find $\sqrt{13}\beta = 2\sqrt{13}$, which simplifies beautifully to $\beta = 2$. We have found the height of our point!

### The Algebraic Bridge

We know the y-coordinate of $P$ is $2$. Since $P$ lies on the hyperbola, its coordinates must satisfy the equation:

$$\frac{\alpha^2}{9} - \frac{\beta^2}{4} = 1$$

Substituting $\beta = 2$ into this equation:

$$\frac{\alpha^2}{9} - \frac{2^2}{4} = 1 \Rightarrow \frac{\alpha^2}{9} - 1 = 1$$

Adding $1$ to both sides, we get $\frac{\alpha^2}{9} = 2$, which means $\alpha^2 = 18$. We have found the square of the x-coordinate!

### The Final Destination

The question asks for the square of the distance of $P$ from the origin. The distance $d$ from the origin $(0,0)$ to $P(\alpha, \beta)$ is given by $d = \sqrt{\alpha^2 + \beta^2}$.

Therefore, $d^2 = \alpha^2 + \beta^2$. We know $\alpha^2 = 18$ and $\beta = 2$, so $\beta^2 = 4$.

Adding these together:

$$d^2 = 18 + 4 = 22$$

And there it is—the result of our journey. The **square of the distance is 22**. It is a testament to the beauty of mathematics that such a complex-sounding problem resolves into such a clean, elegant integer.

Animated Solution for Mathematics - Conic Sections: Let P be a point on the hyperbola $H : \frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , in the first quadrant such that the area of triangle formed by P and the two foci of H is $213$ . Then, the square of the distance of P from the origin is

Select Answer:

Visualized Solution

Standard Equation of Hyperbola

Identifying Parameters $a^{2}$ and $b^{2}$

Calculating Eccentricity $e$

Locating the Foci $S_{1}$ and $S_{2}$

Forming Triangle $P S_{1} S_{2}$

Area of Triangle $P S_{1} S_{2}$

Solving for $y$ -coordinate $β$

Substituting $P$ into Hyperbola

Solving for $α^{2}$

Square of Distance from Origin

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Hyperbola's Hidden Geometry

Decoding the Anatomy

The Geometry of the Triangle

The Algebraic Bridge

The Final Destination

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Select Answer:

Visualized Solution

Standard Equation of Hyperbola

Identifying Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Locating the Foci .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Forming Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS1​S2​

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS1​S2​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P into Hyperbola

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2