Problem 46 - Algebra and Trigonometry

Question

convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$ 9 y^{2}-4 x^{2}-18 y+24 x-63=0 $$

Step-by-Step Solution

Answer

The standard form is \((x - 3)^2/27 - (y + 1)^2/12 = 1\), Foci: (3±√39, -1), Asymptotes: y = ± (2/3)(x - 3) - 1

Step 1: Rearrange the equation

First, group the x and y terms together and switch sides to make it easier to complete the square. Giving us: \[ 4x^2 - 24x + 9y^2 + 18y = 63 \]

Step 2: Complete the squares for x and y

Next, complete the square for each variable by adding the square of half the coefficient of x/y to both sides of equation, which gives us: \[ 4(x^2 - 6x + 9) + 9 (y^2 + 2y + 1) = 63 + 36 + 9 \] The above simplifies to: \[ 4(x - 3)^2 + 9 (y + 1)^2 = 108 \]

Step 3: Convert to standard form

Divide through by 108 to get the hyperbola in the standard form of \((x-h)^2/A^2 - (y-k)^2/B^2 = 1\): \[ (x - 3)^2/27 - (y + 1)^2/12 = 1 \]

Step 4: Compute the values

The center (h,k) is (3,-1), a is \(\sqrt{27}\), b is \(\sqrt{12}\), and \[c = \sqrt{a^2 + b^2} = \sqrt{27 + 12} = \sqrt{39}\]. The foci are therefore (3±√39, -1) and asymptotes are y = ± (b/a)(x - h) + k = ± (2/3)(x - 3) - 1.

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations. The goal is to transform a quadratic expression into a perfect square trinomial.
To achieve this, we first ensure the equation is properly rearranged, grouping the x and y terms.
For the equation \(9y^{2} - 4x^{2} - 18y + 24x - 63 = 0\), we rearrange it to \(4x^2 - 24x + 9y^2 + 18y = 63\).

• Focus on each variable individually: the x terms \(4x^2 - 24x\) and the y terms \(9y^2 + 18y\).
• Find half of the coefficient, square it, and add that value inside the equation to both sides.
• For completeness: \((x^2 - 6x + 9)\) became \((x-3)^2\), and \((y^2+2y+1)\) is \((y+1)^2\).

This step is critical because it allows us to convert the equation into a standard form easily.

Standard Form

For hyperbolas, the standard form resembles \((x-h)^2/A^2 - (y-k)^2/B^2 = 1\). This is a crucial form to understand the geometry of the hyperbola.
Following the completion of the square for our terms, we get \(4(x-3)^2 + 9(y+1)^2 = 108\).

• Divide the entire equation by 108 to normalize the equation into standard form.
• The equation becomes \((x-3)^2/27 - (y+1)^2/12 = 1\).
• Here, \(h\) and \(k\) are the coordinates of the center of the hyperbola, \((h, k) = (3, -1)\).

Knowing the vertices and orientation of both branches allows us to plot and understand the hyperbola's behavior.

Equations of Asymptotes

Asymptotes of a hyperbola are the lines that the hyperbola approaches but never touches.
They are useful for sketching the hyperbola accurately on a graph. For the hyperbola equation \((x-3)^2/27 - (y+1)^2/12 = 1\), the arrangement is critical.

• The formula for the equations of asymptotes is \(y = \pm (b/a)(x-h) + k\).
• Here, \(b/a = \sqrt{12}/\sqrt{27} = 2/3\).
• So, the asymptotes are given by \(y = (\pm 2/3)(x - 3) - 1\).

These lines form a framework around which the hyperbola curves as it branches into its infinite extensions.

Foci of Hyperbolas

The foci are two distinct points that lie inside each open branch of a hyperbola.
Knowing their location helps in defining the hyperbola, as they determine its shape and orientation.
To find the foci, utilize the relationship \(c = \sqrt{a^2 + b^2}\). For our exercise, this is \(c = \sqrt{27 + 12} = \sqrt{39}\).

• The foci are placed at \((h \pm c, k)\), which computes to \((3 \pm \sqrt{39}, -1)\).
• The foci are intrinsic to understanding the geometry of hyperbolas, as the distance from any point on the hyperbola to these foci is constant.

This results in the hyperbola having balanced and perfectly symmetrical branches.