Problem 28 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ y^{2}-3 x^{2}+6 y+6 x-18=0 $$

Step-by-Step Solution

Answer

Center: (1, -3); vertices: (1, -3±2√6); foci: (1, -3±4√2); asymptotes: y+3=±√3(x-1).

Step 1: Rewrite in Standard Form

Start by rewriting the given equation in the standard form of a hyperbola. The equation is \( y^2 - 3x^2 + 6y + 6x - 18 = 0 \). Begin by rearranging terms and grouping the \( y \) and \( x \) terms:\[y^2 + 6y - 3(x^2 - 2x) = 18\]Next, complete the square for both the \( y \) and \( x \) terms.

Step 2: Complete the Square for y

To complete the square for \( y^2 + 6y \):1. Take half of the coefficient of \( y \) (which is 6), square it: \( (\frac{6}{2})^2 = 9 \).2. Add and subtract this square inside the equation:\[(y^2 + 6y + 9) - 9 - 3(x^2 - 2x) = 18\]This simplifies to:\[(y + 3)^2 - 9 - 3(x^2 - 2x) = 18\]

Step 3: Complete the Square for x

For the \( -3(x^2 - 2x) \) term, complete the square:1. Factor out \(-3\) and complete the square inside: \( -3[(x^2 - 2x + 1) - 1] \).2. This becomes \( -3((x-1)^2 - 1) = -3(x-1)^2 + 3 \).Now, rewrite the equation including these squares:\[(y + 3)^2 - 9 - 3(x-1)^2 + 3 = 18\]

Step 4: Simplify and Write the Equation in Standard Form

Simplify the equation from Step 3:\[(y + 3)^2 - 3(x-1)^2 - 6 = 18\]Add 6 to both sides:\[(y + 3)^2 - 3(x-1)^2 = 24\]Divide the entire equation by 24 to get:\[\frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1\]This is the standard form of a hyperbola with its transverse axis vertical.

Step 5: Identify the Vertices and Foci

From the standard form \( \frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1 \):- The center is \((h, k) = (1, -3)\).- \( a^2 = 24 \) (so \( a = \sqrt{24} = 2\sqrt{6} \)).- \( b^2 = 8 \) (so \( b = \sqrt{8} = 2\sqrt{2} \)).- To find \( c \) use \( c^2 = a^2 + b^2 = 24 + 8 = 32 \) \( c = \sqrt{32} = 4\sqrt{2} \).Thus, the vertices are at \((1, -3 \pm 2\sqrt{6})\) and the foci are at \((1, -3 \pm 4\sqrt{2})\).

Step 6: Determine the Asymptotes

For a hyperbola with a vertical transverse axis, the equations of the asymptotes are:\[(y + 3) = \pm \frac{a}{b}(x - 1)\]Substitute \( a = 2\sqrt{6} \) and \( b = 2\sqrt{2} \):\[(y + 3) = \pm \frac{2\sqrt{6}}{2\sqrt{2}} (x - 1)\]This simplifies to:\[(y + 3) = \pm \sqrt{3}(x - 1)\]Therefore, the equations of the asymptotes are \( y + 3 = \sqrt{3}(x - 1) \) and \( y + 3 = -\sqrt{3}(x - 1) \).

Step 7: Graph the Hyperbola

Using the center \((1, -3)\), draw the vertices along the transverse axis (vertical) at \((1, -3 \pm 2\sqrt{6})\). Plot the foci at \((1, -3 \pm 4\sqrt{2})\). Draw the asymptotes lines using their equations. The hyperbola opens upward and downward along the y-axis.

Key Concepts

Vertices

When dealing with a hyperbola, the vertices are key points that define the shape and structure of the curve. They are the points where the hyperbola intersects its transverse axis.. For the equation \( \frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1 \), the center of the hyperbola is located at \((1, -3)\).. Since the transverse axis is vertical, the vertices lie directly above and below this center point. To find them, you calculate using \( a = \sqrt{24} = 2\sqrt{6} \):

• Add \( 2\sqrt{6} \) to the y-coordinate: \( -3 + 2\sqrt{6} \).
• Subtract \( 2\sqrt{6} \) from the y-coordinate: \( -3 - 2\sqrt{6} \).

This results in vertices located at \((1, -3 + 2\sqrt{6})\) and \((1, -3 - 2\sqrt{6})\).. These points are critical when sketching the hyperbola's basic shape, as they mark the closest distance between the branches of the hyperbola along the transverse axis.

Foci

The foci of a hyperbola play a crucial role in its geometric properties. They influence the shape of the hyperbola and are used in its definition.. In the equation \( \frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1 \), the foci are positioned along the same axis as the vertices but further out.. To determine the location of the foci, we calculate the distance \( c \) using the relationship \( c^2 = a^2 + b^2 \):

• Compute \( c = \sqrt{32} = 4\sqrt{2} \).
• Add \( 4\sqrt{2} \) to the y-coordinate of the center: \( -3 + 4\sqrt{2} \).
• Subtract \( 4\sqrt{2} \) from the y-coordinate of the center: \( -3 - 4\sqrt{2} \).

This gives us the foci at \((1, -3 + 4\sqrt{2})\) and \((1, -3 - 4\sqrt{2})\).. The foci are used to construct the hyperbola's curve by ensuring that the difference in distances from any point on the hyperbola to the foci is constant.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never intersects. They help define the overall shape and orientation of the hyperbola.. For a hyperbola with the equation \( \frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1 \), we have vertical asymptotes since the transverse axis is vertical.. The equations of the asymptotes are found using the formula \[ (y + 3) = \pm \frac{a}{b}(x - 1) \] Substituting \( a = 2\sqrt{6} \) and \( b = 2\sqrt{2} \) gives:

• \( (y + 3) = \sqrt{3}(x - 1) \)
• \( (y + 3) = -\sqrt{3}(x - 1) \)

These lines intersect the center at \((1, -3)\) and give the hyperbola its characteristic "X" shape.. The asymptotes effectively guide how the hyperbola's branches open, stretching towards infinity without touching these guiding lines.

Graphing Hyperbolas

Graphing a hyperbola involves several steps but offers a clear visualization of its properties. Starting with the equation \( \frac{(y + 3)^2}{24} - \frac{(x - 1)^2}{8} = 1 \), we can graph by following these points:.. Identify the center, found at \((1, -3)\). This center acts as a reference point for other important features of the hyperbola.
Proceed to plot the vertices, which provide crucial points along the transverse axis: \((1, -3 + 2\sqrt{6})\) and \((1, -3 - 2\sqrt{6})\).
Next, mark the foci at \((1, -3 + 4\sqrt{2})\) and \((1, -3 - 4\sqrt{2})\).. These focus points help in shaping the curve and validating its paths.
Finally, draw the asymptotes from the equations:\( y + 3 = \pm \sqrt{3}(x - 1) \).. These diagonal lines serve as the guides, showing the tendency of the hyperbola as it extends outward.

• Make sure the branches of the hyperbola approach, but never touch, these asymptotes.
• The hyperbola will open upwards and downwards from the center, forming two separate curves as the branches.

By following these steps, you can accurately sketch the hyperbola and understand its geometric relationships.