Problem 8 - College Algebra

Question

Find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: \((-7,0),(7,0) ;\) vertices: \((-5,0),(5,0)\)

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \(x^2/25 - y^2/24 = 1\).

Step 1: Find the Center (h, k)

The center of the hyperbola can be found by finding the midpoint between the two vertices. The midpoint formula is \((x_{1}+x_{2}/2,(y_{1}+y_{2}/2)\). For the vertices \((-5, 0)\) and \((5, 0)\), calculating the midpoint gives us the center, \((0, 0)\). Thus, \(h = 0\) and \(k = 0\).

Step 2: Find the value of 'a'

'a' is the distance from the center of the hyperbola to either vertex. The vertices given are \((-5, 0)\) and \((5, 0)\). Since the center is at \((0, 0)\), the value of 'a' is the x-coordinate of either vertex, so \(a = 5\).

Step 3: Find the value of 'c'

'c' is the distance from the center of the hyperbola to either focus. The foci are given as \((-7, 0)\) and \((7, 0)\). Since the center is at \((0, 0)\), the value of 'c' is the x-coordinate of either focus, so \(c = 7\).

Step 4: Find the value of 'b'

With 'a' and 'c' known, 'b' can be calculated using the formula for a hyperbola \(b = \sqrt{c^2 - a^2}\). Substituting \(c = 7\) and \(a = 5\) into the equation gives: \(b = \sqrt{7^2 - 5^2} = \sqrt{49 - 25} = \sqrt{24}\).

Step 5: Write the equation of the hyperbola

Now with all the variables known, the equation of the hyperbola in standard form can be written as \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\). Substituting \(h = 0\), \(k = 0\), \(a = 5\), \(b = \sqrt{24}\) into the equation gives: \((x-0)^2/5^2 - (y-0)^2/(sqrt{24})^2 = 1\), which simplifies to \(x^2/25 - y^2/24 = 1\).

Key Concepts

Standard Form

The standard form of the equation of a hyperbola is essential for understanding its shape and position in the coordinate plane. The general standard form of a hyperbola centered at

• If the hyperbola opens horizontally, the form is: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]
• If it opens vertically, the form changes to: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]

Here,

• \( (h, k) \) is the center of the hyperbola,
• \( a \) represents the distance from the center to each vertex,
• \( b \) is related to the distance of the co-vertices,
• \( c \) is the distance from the center to each focus.

In the current problem, we calculate the center, vertices, and foci to determine the coefficients \( a, b, \) and \( c \). Therefore, the standard form for the hyperbola with given parameters becomes \( \frac{x^2}{25} - \frac{y^2}{24} = 1 \).

Vertices

Vertices of a hyperbola are points where the hyperbola intersects its transverse axis, and they play a key role in determining the shape and alignment of the hyperbola. Given vertices are

• \((-5, 0)\)
• \((5, 0)\)

From the vertices, we can determine the center

• The center is the midpoint of the vertices, calculated using the midpoint formula.\[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]

Plugging \((-5, 0)\) and \((5, 0)\) into this formula gives the center:

• \((0, 0)\)

The distance from the center to any vertex is \(a\). Thus, \(a = 5\). This tells us the hyperbola extends 5 units left and right from the center along the x-axis.

Foci

The foci (plural for focus) of a hyperbola are critical elements positioned along the transverse axis beyond the vertices. Each hyperbola has two foci. For our hyperbola,

• the coordinates of the foci are given as \((-7, 0)\) and \((7, 0)\).
• The center of the hyperbola is \((0, 0)\).
• The distance from the center to each focus is \(c\).

To find \(c\), measure the distance from the center to one of the foci. In this problem, that distance is

• \(c = 7\),

indicating that the foci are 7 units away from the center. This is essential for crafting the hyperbola equation and understanding its geometry.

Distance Formula

The distance formula helps find the space between two points in a plane and is particularly handy for determining center, vertices, and foci positions when dealing with hyperbolas. It reflects the Pythagorean theorem and is represented as: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Adopting this formula:

• the midpoint distance \[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]provides the center of a hyperbola.

In our context, it was used for calculating the center of the hyperbola, leveraging the given vertices:

• \(d = 5\) for vertices calculation,
• \(d = 7\) for foci calculation.

It also allows us to derive \(b\) through the relation \(c^2 = a^2 + b^2\), where both \(a\) and \(c\) have been determined earlier. Therefore, the distances verify key hyperbola elements and figure into computing all necessary parameters.