Let P(10, 2√15) be a point on the hyperbola x²/a² - y²/b² = 1, whose foci are S and S'. If the lengt

Q: Wait, why do we reject the negative root when solving the quadratic equation for $a$?

That is a fantastic question! In the context of the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the parameter $a$ represents the distance from the center to the vertex along the transverse axis. Since $a$ is a physical distance, it must be strictly positive. If we allowed $a$ to be negative, it would not change the shape of the hyperbola, but it would be mathematically inconsistent with the standard definition where $a > 0$.

Q: Why can't we just use the definition of a hyperbola $|PS - PS'| = 2a$ to find the area?

You absolutely could! The definition $|PS - PS'| = 2a$ is a fundamental property of the hyperbola. However, using that definition would require you to calculate the distances $PS$ and $PS'$ using the distance formula, which involves square roots and can become quite algebraically tedious. The coordinate geometry approach we used is much more direct and less prone to calculation errors in this specific problem.

Q: Is the area of $\Delta PSS'$ always $\frac{1}{2} \times 2c \times y_P$?

Yes, it is! Because the foci $S$ and $S'$ lie on the x-axis, the segment $SS'$ is a horizontal line segment. The base of the triangle is simply the distance between the foci, which is $2c$. The height of the triangle is the perpendicular distance from the point $P$ to the x-axis, which is simply the absolute value of the y-coordinate of $P$. This geometric insight simplifies the area calculation significantly.

Question

Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are $S$ and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :

Options:

(A) 900

(B) 4200

(C) 1462

(D) 2700

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a point $P(10, 2\sqrt{15})$ resting on the hyperbola defined by the equation:

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

Our objective is to determine the square of the area of the triangle formed by this point $P$ and the two foci, $S$ and $S'$.

### The Bridge: The Latus Rectum

Every hyperbola is defined by its latus rectum, a chord passing through a focus perpendicular to the transverse axis. We are given that its length is $8$.

The formula for the length of the latus rectum is:

$$\frac{2b^2}{a} = 8$$

By simplifying this expression, we obtain the relationship $b^2 = 4a$. This equation serves as our bridge, allowing us to express the hyperbola's parameters in terms of a single variable, $a$.

### The Anchor: Point P

Since the point $P(10, 2\sqrt{15})$ lies on the hyperbola, it must satisfy the hyperbola's equation. Substituting the coordinates, we get:

$$\frac{10^2}{a^2} - \frac{(2\sqrt{15})^2}{b^2} = 1 \implies \frac{100}{a^2} - \frac{60}{b^2} = 1$$

By substituting our bridge equation $b^2 = 4a$ into this expression, we arrive at:

$$\frac{100}{a^2} - \frac{60}{4a} = 1 \implies \frac{100}{a^2} - \frac{15}{a} = 1$$

### The Algebra: Solving for $a$

To solve for $a$, we multiply the entire equation by $a^2$ to clear the denominators:

$$100 - 15a = a^2$$

Rearranging this into a standard quadratic form, we get:

$$a^2 + 15a - 100 = 0$$

Factoring the quadratic equation yields $(a + 20)(a - 5) = 0$. This provides two potential values for $a$: $-20$ and $5$. Since $a$ represents a physical distance, it must be positive; therefore, we select **$a = 5$**.

Using our bridge equation, we find $b^2 = 4(5) = 20$.

### The Climax: The Triangle $PSS'$

With $a = 5$ and $b^2 = 20$, we calculate the focal distance $c$ using the fundamental relation $c^2 = a^2 + b^2$:

$$c^2 = 25 + 20 = 45 \implies c = 3\sqrt{5}$$

The foci are located at $S(3\sqrt{5}, 0)$ and $S'(-3\sqrt{5}, 0)$. The base of the triangle $PSS'$ is the distance between these foci:

$$	ext{Base} = 2c = 6\sqrt{5}$$

The height of the triangle is the perpendicular distance from $P$ to the $x$-axis, which is the $y$-coordinate of $P$, $2\sqrt{15}$. The area of the triangle is:

$$	ext{Area} = \frac{1}{2} 	imes (6\sqrt{5}) 	imes (2\sqrt{15}) = 6\sqrt{75} = 30\sqrt{3}$$

The problem asks for the square of this area:

$$(30\sqrt{3})^2 = 900 	imes 3 = 2700$$

The final result is **$2700$**.

Animated Solution for Mathematics - Conic Sections: Let $P (10, 215)$ be a point on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , whose foci are $S$ and $S^{'}$ . If the length of its latus rectum is 8, then the square of the area of $Δ P S S^{'}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Point $P$

The Latus Rectum Constraint

Substituting Point $P$

Merging the Equations

Forming the Quadratic Equation

Solving for Parameter $a$

Finding $b^{2}$ and Focal Distance $c$

Locating the Foci and Triangle

Dimensions of $Δ P S S^{'}$

Calculating the Area

Final Answer: Square of the Area

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

Analyzing the Setup

The Bridge

The Anchor

The Algebra

The Climax

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