Problem 13 - Algebra 2

Question

Find the foci of each hyperbola. Then draw the graph. $$ \frac{x^{2}}{64}-\frac{y^{2}}{36}=1 $$

Step-by-Step Solution

Answer

The foci of the hyperbola \( \frac{x^{2}}{64}-\frac{y^{2}}{36}=1\) are located at \((±10, 0)\).

Step 1: Compute the semi-major and semi-minor axes

The lengths of the semi-major axis \(a\) and semi-minor axis \(b\) can be found by taking the square roots of their squared values. Here, for the given equation, we have \(a^2 = 64\) and \(b^2 = 36\), which gives \(a = 8\) and \(b = 6\).

Step 2: Calculate the foci

The foci of a hyperbola are located at \(± \sqrt{a^2 + b^2}\) on the x-axis. So, calculate the foci using this formula. This results in the foci at \((± \sqrt{64+36}, 0) = (±10, 0)\).

Step 3: Draw the graph

To draw the graph of the hyperbola, start by drawing the x and y axes. Place the center of the hyperbola at the origin. Sketch the two branches of the hyperbola, centered at the origin, and extending out to pass through the vertices at \((±a, 0)\), which are \((±8, 0)\) for this problem. Indicate the location of the foci along the x-axis at \((±10, 0)\). The hyperbola opens along the x-axis with the foci at \((±10, 0)\).

Key Concepts

Foci of Hyperbola

When diving into the world of hyperbolas, one of the unique features to understand is the foci, which are two distinct points located along the main axis of this beautiful curve. These points play a critical role in defining the shape and nature of the hyperbola.
For the equation \[ \frac{x^{2}}{64} - \frac{y^{2}}{36} = 1 \],we can identify the foci using a specific formula. The foci for hyperbolas are calculated as:

• \( \pm \sqrt{a^2 + b^2} \)

Here, given \( a^2 = 64 \) and \( b^2 = 36 \), we calculate \( \sqrt{64+36} = 10 \).
This means our foci are positioned at \((±10, 0)\) along the x-axis. The presence and location of the foci help determine the internal geometry of the hyperbola, guiding the shape as the curve approaches them but never crosses through.

Semi-Major and Semi-Minor Axes

Understanding the semi-major and semi-minor axes is crucial when studying hyperbolas, as these axes essentially dictate the stretch and orientation of the hyperbola's branches.
The semi-major axis is the longest distance from the center to the vertices along the main axis of the hyperbola, while the semi-minor axis stretches perpendicularly to it.

• For our equation: \( \frac{x^{2}}{64} - \frac{y^{2}}{36} = 1 \), we find the semi-major axis is calculated as \( a = \sqrt{64} = 8 \)
• The semi-minor axis comes out to be \( b = \sqrt{36} = 6 \)

The lengths of these axes inform us that the hyperbola extends horizontally in this scenario due to a being larger than b. It helps to visualize the path the hyperbola will take and assists in identifying how far each branch will expand along the x-axis compared to the y-axis.

Graphing Hyperbolas

Graphing a hyperbola can at first seem daunting, but with a systematic approach, it becomes manageable and insightful. Begin by identifying the center of the hyperbola, which is typically at the origin if the equation is centered like in our case:
\[ \frac{x^{2}}{64} - \frac{y^{2}}{36} = 1 \].
Next, plot the vertices of the hyperbola using the lengths of the semi-major and semi-minor axes.
For this equation, the vertices are located at:

• \((±8, 0)\) along the x-axis

Now, mark the foci previously calculated at \((±10, 0)\).
The hyperbola's branches will open outward from these vertices along the x-axis, curving towards the foci but never meeting them. Visualize how the curve will pass through these key points and sketch the asymptotes, which in standard form intersect through the center intersecting points. Hyperbolas open along the axis where the larger denominator is located, so here the curve stretches horizontally, reflecting the x-axis dominance in our equation.Watching as these elements form together gives life to the hyperbola, making it an engaging topic in analytical geometry.