Problem 27 - Algebra 2

Question

Graph each equation. $$ 5 x^{2}-12 y^{2}=120 $$

Step-by-Step Solution

Answer

The hyperbola equation \(5x^{2}-12y^{2}=120\) has its center at (0,0), vertices at (\(\sqrt{24}\),0) and (-\(\sqrt{24}\),0), foci at (\(\sqrt{34}\), 0) and (-\(\sqrt{34}\), 0), and asymptotes with the equations \( y= \pm\frac{\sqrt{15}}{12}x \).

Step 1: Identify the Centre

The equation does not have terms involving x and y, therefore, the center of the hyperbola would be at (0,0).

Step 2: Find the semi-axes

Rewrite the given equation \(5x^{2}-12y^{2}=120\) in standard form. We have \(\frac{x^{2}}{24} - \frac{y^{2}}{10} = 1\). Thus the semi-major axis is \(a = \sqrt{24}\) and the semi-minor axis is \(b = \sqrt{10}\)

Step 3: Determine the vertices and foci

The vertices are at a distance of 'a' from the center in the x direction since x-axis term has the positive sign in the equation. Therefore, the vertices are at (\(\sqrt{24}\), 0) and (-\(\sqrt{24}\), 0). The foci are at a distance of \(c=\sqrt{a^{2}+b^{2}} = \sqrt{24 + 10} = \sqrt{34}\) from the center. Hence, the foci will be at (\(\sqrt{34}\), 0) and (-\(\sqrt{34}\), 0).

Step 4: Find the asymptotes

The equations for the asymptotes of the hyperbola are \( y= \pm\frac{b}{a}x = \pm\frac{\sqrt{10}}{\sqrt{24}}x = \pm \frac{\sqrt{15}}{12}x \). These lines pass through the center of the hyperbola and have slopes of \(\pm\frac{b}{a}\).

Step 5: Sketch the graph of the hyperbola

Draw the hyperbola with the center at (0,0), vertices at (\(\sqrt{24}\),0) and (-\(\sqrt{24}\),0), foci at (\(\sqrt{34}\), 0) and (-\(\sqrt{34}\), 0), and asymptotes which have the equations \( y= \pm\frac{\sqrt{15}}{12}x \).

Key Concepts

Equation of a Hyperbola

A hyperbola is a type of conic section that is formed when a plane intersects a double-napped cone at an angle. Unlike ellipses and circles, hyperbolas open in opposite directions. The standard form for the equation of a hyperbola that opens horizontally is given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Meanwhile, for a vertically opening hyperbola, the equation becomes \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

In this exercise, the equation is originally \( 5x^2 - 12y^2 = 120 \). To bring it to the standard form, we rewrite it as \( \frac{x^2}{24} - \frac{y^2}{10} = 1 \). Here, \( a = \sqrt{24} \) and \( b = \sqrt{10} \) as we can see from the denominators. Knowing the standard form helps in identifying various components of the hyperbola such as vertices, foci, and asymptotes.

Vertices of a Hyperbola

Vertices are crucial in defining the shape and direction of a hyperbola. For a hyperbola centered at the origin, the vertices are located at a distance 'a' along the transverse axis. If the hyperbola opens horizontally, this axis is aligned with the x-axis. If it opens vertically, it aligns with the y-axis. Simply put:

• Horizontally oriented: Vertices are at \((\pm{a}, 0)\).
• Vertically oriented: Vertices are at \((0, \pm{a})\).

The equation \( \frac{x^2}{24} - \frac{y^2}{10} = 1 \) indicates a horizontally oriented hyperbola as the \( x^2 \) term is positive. Therefore, the vertices are at \( (\sqrt{24}, 0) \) and \( (-\sqrt{24}, 0) \). This shows that vertices lie along the x-axis.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points located along the transverse axis, which help in defining the hyperbola's precise shape. The distance from the center to each focus is always more significant than the distance to a vertex. You can calculate this distance using the formula \( c = \sqrt{a^2 + b^2} \).

For this particular problem, we determined that \( a = \sqrt{24} \) and \( b = \sqrt{10} \). Therefore, the distance 'c' to the foci from the center is \( c = \sqrt{24 + 10} = \sqrt{34} \). Since our hyperbola opens horizontally, the foci are located at \( (\sqrt{34}, 0) \) and \( (-\sqrt{34}, 0) \). The foci are always aligned along with the transverse axis.

Asymptotes of a Hyperbola

Asymptotes are imaginary lines that a hyperbola approaches but never intersects. They provide a useful guideline to sketching the general location and shape of a hyperbola on a graph. Asymptotes for hyperbolas of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) have equations given by \( y = \pm \frac{b}{a} x \).

Applying the given equation \( \frac{x^2}{24} - \frac{y^2}{10} = 1 \), we use \( a = \sqrt{24} \) and \( b = \sqrt{10} \) to find the asymptote equations: \( y = \pm \frac{\sqrt{10}}{\sqrt{24}} x \) or simplified \( y = \pm \frac{\sqrt{15}}{12} x \). These asymptotes intersect at the center of the hyperbola, creating a perfect framework for sketching it accurately. They visually illustrate how the branches of the hyperbola open and diverge, simplifying graph drawing immensely.