Problem 62 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ 9 x^{2}+6 y^{2}+36 y=0 $$

Step-by-Step Solution

Answer

The center is (0, -3), vertices are at (±√6, -3), foci are at (±√15, -3), and the asymptotes are y = -3 ± (√6/2)x.

Step 1: Rewrite the Equation

First, rewrite the given equation to better identify the conic section. The equation given is \(9x^2 + 6y^2 + 36y = 0\). Observe that this equation involves both \(x^2\) and \(y^2\) terms, indicating it may be an ellipse or a hyperbola. Begin by restructuring the equation: move \(y\)-related terms to one side: \(9x^2 = -6y^2 - 36y\).

Step 2: Complete the Square

To simplify the equation, complete the square for the \(y\) terms. Factor out \(-6\) from \(-6y^2 - 36y\) to get \(-6(y^2 + 6y)\). Now complete the square: \(y^2 + 6y\) becomes \((y+3)^2 - 9\). The equation now reads \(9x^2 = -6((y+3)^2 - 9)\).

Step 3: Simplify the Equation

Distribute \(-6\) into the squared expression: \(9x^2 = -6(y+3)^2 + 54\). Rearrange to standard form: \(9x^2 - 6(y+3)^2 = -54\). By dividing the whole equation by \(-54\), we'll get \( \frac{x^2}{6} - \frac{(y+3)^2}{9} = 1\). This represents a hyperbola.

Step 4: Identify the Center

For the hyperbola \( \frac{x^2}{6} - \frac{(y+3)^2}{9} = 1\), the general form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). By comparing, the center \((h, k)\) is \( (0, -3) \).

Step 5: Identify Vertices

The vertices of a hyperbola in the form \( \frac{x^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) are located at \((h+a, k)\) and \((h-a, k)\). Thus, the vertices are at \((\sqrt{6}, -3)\) and \((-\sqrt{6}, -3)\).

Step 6: Identify Foci

The foci of a hyperbola are given by \((h + c, k)\) and \((h - c, k)\), where \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{6 + 9} = \sqrt{15}\), so the foci are at \((\sqrt{15}, -3)\) and \((-\sqrt{15}, -3)\).

Step 7: Find Asymptotes

The asymptotes for this hyperbola are \(y = k \pm \frac{b}{a}(x)\). Here, \(k = -3\), \(b = 3\), and \(a = \sqrt{6}\), so the equations for the asymptotes are \(y = -3 \pm \frac{3}{\sqrt{6}}x\), which simplifies to \(y = -3 \pm \frac{\sqrt{6}}{2}x\).

Key Concepts

Hyperbola

A hyperbola is a type of conic section—it is a curve that you get when cutting a cone with a plane at a steep angle. Hyperbolas are characterized by their two disjoint curves. If you think of a hyperbola like a pair of mirrored arcs, it may help! The standard form of a hyperbola can be written as \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) or \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \). The choice of equation depends on the orientation of the hyperbola, whether it opens horizontally or vertically.

• If the \(x\) term comes first, the hyperbola is horizontal.
• If the \(y\) term comes first, the hyperbola is vertical.

Understanding this basic structure, as in our solution where the equation becomes \( \frac{x^2}{6} - \frac{(y+3)^2}{9} = 1 \), helps you identify a hyperbola and distinguish it from other conic sections, like ellipses.

Vertices

Vertices are key points on a hyperbola that help define its shape. They are located at the points where the hyperbola intersects with its axis. To find the vertices' positions, we use the equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).

To determine the coordinates:

• For a horizontal hyperbola: The vertices lie at \((h \pm a, k)\).
• For a vertical hyperbola: The vertices are at \((h, k \pm b)\).

In the exercise, our hyperbola's vertices at \((\sqrt{6}, -3)\) and \(( -\sqrt{6}, -3)\) tell us it is horizontal because the changes occur along the \(x\)-axis, defined by \(a = \sqrt{6}\).

Foci

The foci of a hyperbola are two specific points that lie along the axis of the hyperbola, further from the center than its vertices. They play a crucial role in the geometric definition of a hyperbola. For a hyperbola, the sum of the distances from any point on the hyperbola to the two foci is constant.

The formula to find the foci utilizes \(c\), which is calculated using \(c = \sqrt{a^2 + b^2} \). Once \(c\) is determined:

• For a horizontal hyperbola: The foci are at \((h \pm c, k)\).
• For a vertical hyperbola: The foci are located at \((h, k \pm c)\).

In this problem, calculating \(c\) gives \(c = \sqrt{15}\), placing our foci at \((\sqrt{15}, -3)\) and \((-\sqrt{15}, -3)\), reflecting how they sit outside the vertices along the \(x\)-axis.

Asymptotes

Asymptotes are lines that the branches of a hyperbola approach but never touch. They serve as important guides for sketching the hyperbola's shape. Unlike finite parts of the graph, they extend indefinitely.

The equations of the asymptotes for a hyperbola are derived from its standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) and are given by:

• \( y = k \pm \frac{b}{a}(x-h) \) for hyperbolas with horizontal branches.
• \( y = k \pm \frac{a}{b}(x-h) \) for hyperbolas with vertical branches.

This equation makes the asymptotes appear as lines cutting through the center of the hyperbola.

In our example, using \( b = 3 \) and \( a = \sqrt{6} \), the asymptotes are \( y = -3 \pm \frac{3}{\sqrt{6}}x \), which simplifies to \( y = -3 \pm \frac{\sqrt{6}}{2}x \). This carefully tells you where the hyperbola's arms spread off to infinity.