Problem 50 - Algebra 2

Question

Which hyperbola does NOT have \((0, \pm 4)\) as its \(y\) -intercepts? $$ \begin{array}{ll}{\text { F. } y^{2}-x^{2}=16} & {\text { G. } 4 y^{2}-16 x^{2}=64} \\ {\text { H. } \frac{x^{2}}{25}-\frac{y^{2}}{16}=1} & {\text { J. } \frac{y^{2}}{16}-\frac{x^{2}}{9}=1}\end{array} $$

Step-by-Step Solution

Answer

The hyperbola H, represented by equation \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\) does not have \((0, ±4)\) as its \(y\) -intercepts.

Step 1: Substitute x=0 in option F

For equation F, we put \(x = 0\). The equation is \(y^{2}-x^{2}=16\). After substitution, it turns out to be \(y^{2}=16\). The solutions for \(y\) are \(±4\). Hence, option F is not the answer.

Step 2: Substitute x=0 in option G

For equation G, we put \(x = 0\). The equation is \(4y^{2}-16x^{2}=64\). After substitution, it turns out to be \(4y^{2}=64\). Dividing by 4, we get \(y^{2}=16\) and the solutions for \(y\) are \(±4\). Hence, option G is not the answer.

Step 3: Substitute x=0 in option H

For equation H, we put \(x = 0\). The equation is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\). After substitution, it turns out to be \(-\frac{y^{2}}{16}=1\). Hence, option H is the answer because it does not yield \(y = ±4\). There is no need to check option J.

Key Concepts

Understanding Y-Intercepts in Hyperbolas

In the context of hyperbolas, y-intercepts are points where the hyperbola intersects the y-axis. To find these points, we substitute \(x = 0\) in the equation of the hyperbola and solve for \(y\). For example, consider the hyperbola represented by the equation \(y^{2} - x^{2} = 16\).

By setting \(x = 0\), the equation simplifies to \(y^{2} = 16\). Solving it, we get \(y = \pm 4\). Hence, the y-intercepts are at \((0, 4)\) and \((0, -4)\).

This process is crucial to determine if specific points, like \((0, \pm 4)\), are the y-intercepts of given hyperbolas. It's essential to check each hyperbola's equation to confirm if such points align with the resulting y-values.

Equations of Conics

Conic sections, like hyperbolas, are expressed in specific mathematical equations. Hyperbolas are generally given by the equation \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\) or \(\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1\). Each form has a distinct orientation and properties.

The equation \(y^{2} - x^{2} = 16\) can be rearranged into the standard form of a hyperbola. This standard equation helps determine features like the center, vertices, and axes of the hyperbola.

• If \(y^{2}\) is first, the hyperbola opens up and down.
• If \(x^{2}\) is first, it opens left and right.

These characteristics are integral for graphing and understanding the hyperbola's behavior on coordinate planes.

Solution of Equations for Identifying Hyperbolas

Solving equations is fundamental in identifying and classifying hyperbolas. In given problems, testing each equation can reveal critical properties, such as y-intercepts. By substituting \(x = 0\), as seen in other examples, we deduce whether a specified y-intercept exists for the hyperbola.

For instance, take \(4y^{2} - 16x^{2} = 64\). By setting \(x = 0\), it becomes \(4y^{2} = 64\). Simplifying gives \(y^{2} = 16\), resulting in \(y = \pm 4\). This matches the desired y-intercepts. Solving like this validates certain options while eliminating others, such as determining the correct or incorrect equation that doesn't support given points.

Role of Coordinate Geometry in Analyzing Hyperbolas

Coordinate geometry allows us to visualize and solve problems related to hyperbolas and other conics. By using coordinate planes, we identify important features like intercepts, axes, and vertices of hyperbolas from their equations.

When analyzing a hyperbola such as \(\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1\), coordinate geometry helps position the hyperbola on the plane, emphasizing where it might intersect axes. Graphing these equations visually aids in seeing how transformations, like translations or rotations, affect the conic's position and orientation.

It is an essential tool for not just solving equations but also for providing insightful visual representations that further increase our understanding of these mathematical forms.