Problem 49 - Algebra 2

Question

Which hyperbola has \(( \pm 6,0)\) as its \(x\) -intercepts? A. \(y^{2}-x^{2}=36 \quad\) B. \(\frac{y^{2}}{36}-\frac{x^{2}}{49}=1\) C. \(\frac{x^{2}}{25}-\frac{y^{2}}{36}=1 \quad\) D. \(\frac{x^{2}}{36}-\frac{y^{2}}{4}=1\)

Step-by-Step Solution

Answer

Answer is D. \(\frac{x^{2}}{36}-\frac{y^{2}}{4}=1\)

Step 1: Apply the Definition of x-intercepts

x-intercepts are points where y = 0. Hence set y = 0 in each of the given options and solve the resulting equations for x.

Step 2: Test Option A

Option A reads \(y^{2}-x^{2}=36\). Substituting y = 0 gives: \(-x^{2}=36\), or equivalently \(x^{2}=-36\). This does not yield real solutions.

Step 3: Test Option B

Option B reads \(\frac{y^{2}}{36}-\frac{x^{2}}{49}=1\). Substituting y = 0 and rearranging, we get: \(\frac{x^{2}}{49}=1\), or \(x^{2}=49\). The solutions are \(x=\pm 7\), which doesn't match our required x-intercepts.

Step 4: Test Option C

Option C reads \(\frac{x^{2}}{25}-\frac{y^{2}}{36}=1\). Substituting y = 0 gives us: \(\frac{x^{2}}{25}=1\), or \(x^{2}=25\). The solutions are \(x=\pm 5\), which doesn't match our required x-intercepts.

Step 5: Test Option D

Option D reads \(\frac{x^{2}}{36}-\frac{y^{2}}{4}=1\). Substituting y = 0 gives us: \(\frac{x^{2}}{36}=1\), or \(x^{2}=36\). The solutions are \(x=\pm 6\), which matches our required x-intercepts.

Key Concepts

x-intercepts

Understanding the concept of x-intercepts is essential to solving problems involving hyperbolas. An x-intercept occurs where the graph of a function crosses the x-axis. This is where the y-value of the function is zero.
When solving for x-intercepts, set the equation of the hyperbola so that \( y = 0 \). This reduces the complexity and allows one to solve for x directly.
The x-intercepts provide critical information about the graph's intersection points along the x-axis. In the given exercise, the values \( (\pm 6, 0) \) were specified as the desired x-intercepts. This guides us toward the correct hyperbola equation, by eliminating the ones that do not satisfy this condition.

solving equations

To find the x-intercepts of a hyperbola, solve the equation step-by-step with substitution. First, substitute \( y = 0 \) into the hyperbola's equation. This simplifies the problem to a basic algebraic equation where x is the sole variable.

• Check each equation to ensure it yields real solutions for x.
• Ensure the calculated solutions match the required x-intercepts.

For instance, in Option D: When \( y = 0 \), the equation becomes \( \frac{x^2}{36} = 1 \). Solving, we find \( x^2 = 36 \), leading to \( x = \pm 6 \). This matches our required x-intercepts, confirming Option D as correct. Accuracy in these calculations is crucial as they lead to identifying the right equation.

quadric surfaces

Quadric surfaces include curves such as hyperbolas, ellipses, and parabolas. Each type has unique geometric properties and equations.
Hyperbolas often appear as two disconnected "mirror" arcs. The standard equation can be represented as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or it can be expressed for vertical orientation.
Determining the specific characteristics of a hyperbola—like its x-intercepts—helps in understanding the surface's shape and orientation.

• An equation modifies its inclusion or orientation depending on how the x and y terms are structured.
• One can express the quadric surface equation differently based on requirements, such as focusing on x-intercepts in the original exercise.

In exercises like the given one, recognizing these properties and relationships between variables and terms is a vital aspect of both graph interpretation and solution of quadric surface equations.