Problem 32 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and asymptotes of the hyperbola, and (c) sketch the hyperbola. Use a graphing utility to verify your graph. \(25 x^{2}-4 y^{2}=100\)

Step-by-Step Solution

Answer

Standard form: \((x^2)/4 - (y^2)/25 = 1\), center: (0, 0), vertices: (2, 0) and (-2, 0), foci: (√29, 0) and (-√29, 0), asymptotes: y = 5/2x and y = -5/2x.

Step 1: Convert to Standard Form

First, make sure the equation is in standard form. The standard form of the equation of a hyperbola is \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2} = 1\). Do this by dividing \(25x^{2}-4y^{2}=100\) by 100 to get the equation in standard form, which is \((x^2)/4 - (y^2)/25 = 1\).

Step 2: Identify the center, vertices, and foci

From the equation, the center of the hyperbola is (h, k) = (0, 0) because there are no shifts in the x or y direction. The values of a and b can be found by taking the square root of the denominators of the x and y terms, respectively. Hence a = 2 and b = 5. \nThe vertices are found at (±a, 0), so the vertices are at (±2, 0). \nNext, we need to find the foci. The distance from the center to the foci is c, which can be found by the equation \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{4 + 25} = \sqrt{29}\). The foci are at (0+±c, 0), or (±√29, 0)

Step 3: Find the Asymptotes

The asymptotes of the hyperbola are given by the equations y = ±b/a * (x-h) + k. Here, the equations of the asymptotes are y = ±5/2 * x, so the asymptotes are y = ±5/2 x.

Step 4: Sketch the Hyperbola

Finally, sketch the hyperbola. Plot the center, vertices, and foci. Sketch in the asymptotes as diagonal lines. The hyperbola approaches these lines as x approaches ± infinity.

Key Concepts

Standard Form of Hyperbola

To find the standard form of a hyperbola's equation, it is essential to transform the given equation into a specific structure. For hyperbolas, the standard form can be identified as \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] for a horizontal hyperbola. The equation \[25x^2 - 4y^2 = 100\] needs conversion. Start by dividing each term by 100, leading to \[\frac{x^2}{4} - \frac{y^2}{25} = 1.\] Now, the equation is in the standard form, revealing key components such as denominators that will be used to find other properties of the hyperbola.
This transformation uncovers that it is a horizontal hyperbola since the x-term is positive.

Center of Hyperbola

The center of a hyperbola is its midpoint, where the two axes intersect. In the standard form equation, \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,\] "h" and "k" represent its coordinates \((h, k)\). In our example, the transformed equation \[\frac{x^2}{4} - \frac{y^2}{25} = 1\] contains no \((x-h)^2\) or \((y-k)^2\) involving subtraction. This indicates no shift from the origin, meaning the center is at \((0, 0)\).
The center is a pivotal component, serving as the reference point to locate other features like vertices and foci.

Vertices of Hyperbola

The vertices of a hyperbola are the points where it intersects the transverse axis (the x-axis for horizontal hyperbolas). To find these points, rely on the equation's denominator of the x-term: \((x-h)^2/a^2\) gives us "a" by taking the square root of 4. Here, \(a = 2\).
The vertices are symmetric around the center, lying at \(\pm a\) units away in the x-direction. In our hyperbola centered at \((0, 0)\), the vertices are located at:

• (2, 0)
• (-2, 0)

These points serve as bounds for the hyperbola's graphed curve along the transverse axis.

Foci of Hyperbola

The foci of a hyperbola are crucial as these points help define its overall shape and direction. They are located along the transverse axis, further from the center than the vertices. To find them, use the formula:\[c = \sqrt{a^2 + b^2}.\]With \(a = 2\) and \(b = 5\), compute:\[c = \sqrt{4 + 25} = \sqrt{29}.\]Placing these coordinates on our graph at

• \((\sqrt{29}, 0)\)
• \((-\sqrt{29}, 0)\)

They are pivotal in the accurate graphing of the hyperbola since the shape extends toward these foci at infinity.
They essentially "guide" the branches as they project away from the center.

Asymptotes of Hyperbola

Asymptotes of a hyperbola serve as imaginary boundary lines that the curve of the hyperbola approaches but never intersects. For our hyperbola, the asymptotes are derived from the formula:\[y = \pm \frac{b}{a}(x-h) + k.\]Here, replace "a" and "b" with their values: \(a = 2\) and \(b = 5\), while \((h, k)\) is the center \((0,0)\). Thus, the equations become:

• \(y = \frac{5}{2}x\)
• \(y = -\frac{5}{2}x\)

These lines, when graphed, crisscross through the center, helping sketch the hyperbola accurately.
The branches of the hyperbola hug closer to these lines as they move outward, making them an integral part of sketching any hyperbola.