Problem 31 - Contemporary Precalculus

Question

Identify the conic whose equation is given and find its graph. If it is an ellipse, list its center, vertices, and foci. If it is a hyperbola, list its center, vertices, foci, and asymptotes. $$4 y^{2}-x^{2}+6 x-24 y+11=0$$

Step-by-Step Solution

Answer

The given equation represents a hyperbola with the center at (3, 3), vertices at (3+√20, 3) and (3-√20, 3), foci at (3+√25, 3) and (3-√25, 3), and asymptotes with slopes ±(√5/√20).

Step 1: Rewrite the equation in standard form

To rewrite the given equation in standard form, we will first group the x-terms and the y-terms, and then complete the square for each group: $$-x^2 + 6x + 4y^2 - 24y = -11$$ Now complete the square for each group: Add \((\frac{1}{2}\cdot6)^2=9\) to both sides for the x-group and add \((\frac{1}{2}\cdot\frac{-24}{4})^2=36\) to both sides for the y-group: $$(-x^2 + 6x + 9) + 4(y^2 - 6y + 9) = -11 + 9 + 4\cdot9$$ Now rewrite the equation in standard form: $$-(x-3)^2 + 4(y-3)^2 = 20$$ Divide both sides by 20: $$\frac{(x-3)^2}{20} - \frac{(y-3)^2}{5} = 1$$

Step 2: Identify the conic section

Observe that the equation is of the form $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$ which corresponds to a hyperbola.

Step 3: Find the center, vertices, and foci

For a hyperbola, we can find the center, vertices, foci, and asymptotes using the following formulas: Center: \((h, k) = (3, 3)\) Vertices: \((h\pm a, k) = (3\pm\sqrt{20}, 3)\) Foci: \((h\pm\sqrt{a^2+b^2}, k) = (3\pm\sqrt{20 + 5}, 3)\) Asymptotes: lines passing through the center with slope \(\pm\frac{b}{a} = \pm\frac{\sqrt{5}}{\sqrt{20}}\)

Step 4: Write the final answer

The equation represents a hyperbola with center \((3, 3)\), vertices \((3+\sqrt{20}, 3)\) and \((3-\sqrt{20}, 3)\), foci \((3+\sqrt{25}, 3)\) and \((3-\sqrt{25}, 3)\), and asymptotes with slopes \(\pm\frac{\sqrt{5}}{\sqrt{20}}\).

Key Concepts

Hyperbola

A hyperbola is a type of conic section, which means it is one of the shapes you get when you slice a cone with a plane. It is characterized by two separate curves called branches. The standard form of a hyperbola equation is important to identify its various features. For instance, the equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) describes a hyperbola with horizontal transverse axis. This equation not only helps in identifying the conic as a hyperbola, but also aids in determining its center, vertices, and other properties.

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This method is essential when rewriting the equation of a conic, like a hyperbola, in its standard form. For instance, given the quadratic terms, you adjust each group by adding and subtracting the square of half the linear coefficient. This is done for both x and y terms in the equation. In the context of the given problem, by completing the square, the expression \((-x^2 + 6x)\) is transformed into \(-(x-3)^2\), and similarly, \((4y^2 - 24y)\) becomes \(4(y-3)^2\). This step is crucial in identifying the conic and subsequently determining its center and vertices.

Center and Vertices of Conics

The center of a hyperbola is the midpoint between its two foci. It is where the symmetry of the hyperbola is centered around. By completing the square, one can find the coordinates of this center. For this hyperbola, the center \(h, k\) comes out to be at point (3, 3).
The vertices are the points where the hyperbola intersects its transverse axis—the main line passing through both branches. With a hyperbola’s standard form equation, these vertices can be calculated using the formula (h\(\pm a\), k) for a horizontal hyperbola. Hence, for the given hyperbola, the vertices are \(3\pm\sqrt{20}, 3\). Calculating these points aids in sketching the graph of the hyperbola accurately.

Foci and Asymptotes of Hyperbolas

The foci of a hyperbola are special points inside each branch that are used to define the shape. The importance of these points lies in their role in the hyperbola's definition: the absolute difference in distances from any point on the hyperbola to the two foci is constant. Using the formula \(h\pm\sqrt{a^2+b^2}, k\), foci for the given hyperbola are found to be at the points (3\(\pm\sqrt{25}\), 3).
Asymptotes are lines that the hyperbola approaches but never touches. They provide a visual guide that indicates the general direction in which each branch of the hyperbola extends. The slopes of these lines are determined by the formula \( \pm \frac{b}{a} \), giving slopes of \( \pm \frac{\sqrt{5}}{\sqrt{20}} \). Drawing these asymptotes on any hyperbola helps in visualizing its graphical behavior as it extends towards infinity.