Problem 47 - 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. $$ 4 x^{2}-9 y^{2}-16 x+54 y-101=0 $$

Step-by-Step Solution

Answer

The equation in standard form is \((x-2)^{2}/11 - (y-3)^{2}/4 = 1\). The foci are \((2- sqrt(15), 3)\) and \((2+ sqrt(15), 3)\). The equations of the asymptotes are \(y = 3 + (2/sqrt(11))(x - 2)\) and \(y = 3 - (2/sqrt(11))(x - 2)\)

Step 1: Rearrange and group

First, the original equation needs to be rearranged and grouped for completing the square. So, group \(x\) terms, \(y\) terms and constants: \(4 x^{2}-16 x -9 y^{2}+54 y = 101\).

Step 2: Complete the square

Next, each group is completed by adding the square of half the coefficient of \(x\) or \(y\) to both sides of the equation. The equation becomes: \(4(x^{2}-4x+4)-9(y^{2}-6y+9)=101 + 4*4 - 9*9 = 44\). Simplifying, we get: \(4(x-2)^{2}-9(y-3)^{2}=44\)

Step 3: Standard form

Divide both sides by 44 to the standard form of a hyperbola. This gives us: \((x-2)^{2}/11 - (y-3)^{2}/4 = 1\)

Step 4: Find Center, a, b

From the standard form, the center of the hyperbola is (2, 3). Further, \(a = sqrt(11)\) and \(b = 2\).

Step 5: Find c & Foci

The distance from the center to the foci is \(c\), which satisfies \(c^{2}=a^{2}+b^{2}\). So, \(c=sqrt(a^{2}+b^{2}) = sqrt(15)\). The foci will be \((2- sqrt(15), 3)\) and \((2+ sqrt(15), 3)\)

Step 6: Find Equations of Asymptotes

The equations of the asymptotes to a hyperbola with center \((h, k)\), semi-major axis length \(a\) and semi-minor axis length \(b\) are given by \(y-k = ±(b/a)*(x-h)\). Therefore, the slopes of the asymptotes are given by \(±2/sqrt(11)\) and the equations of asymptotes are \(y = 3 + (2/sqrt(11))(x - 2)\) and \(y = 3 - (2/sqrt(11))(x - 2)\)

Key Concepts

Completing the Square

Completing the square is a method used to make a quadratic expression into a perfect square trinomial. This technique is useful in various topics, including rewriting equations of hyperbolas in standard form. In this exercise, the equation \(4x^{2} - 16x - 9y^{2} + 54y = 101\) needs completing the square for both \(x\) and \(y\). To do this, you group the \(x\) terms and \(y\) terms separately:

• For \(x\): \(4(x^{2} - 4x)\)
• For \(y\): \(-9(y^{2} - 6y)\)

Then, you find half the coefficient of the \(x\) or \(y\) term, square it, and add it inside the appropriate parentheses, compensating by adjusting the other side of the equation:

• For \(x\): Half of \(-4\) is \(-2\); \((-2)^{2} = 4\).
• For \(y\): Half of \(-6\) is \(-3\); \((-3)^{2} = 9\).

Completing the square transforms the equation to: \[4(x - 2)^{2} - 9(y - 3)^{2} = 44\] This method makes it easier to convert the quadratic equation to its standard form.

Standard Form

The standard form of a hyperbola is a way of expressing its equation to highlight key features such as its center and orientation. Once you have completed the square, the next step is to divide the entire equation by the constant on the right side to set it to 1. For this hyperbola, after completing the square, the equation becomes: \[4(x - 2)^{2} - 9(y - 3)^{2} = 44\] To convert it to standard form, divide everything by 44. This gives: \[ \frac{(x - 2)^{2}}{11} - \frac{(y - 3)^{2}}{4} = 1 \] In this form, you identify:

• The center of the hyperbola at \((h, k) = (2, 3)\).
• \(a^{2} = 11\) and \(b^{2} = 4\), indicating the distances between the center and vertices along the transverse and conjugate axes.

This form makes it easier to identify and graph the hyperbola and find other important features, such as the foci and asymptotes.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never actually touches. They provide the general direction in which the hyperbola opens. These lines are essential for graphing the hyperbola. For a hyperbola with its standard form \(\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1\), the asymptotes can be found using the formula:\[y - k = \pm \frac{b}{a}(x - h)\] Given our example with \(a^{2} = 11\) and \(b^{2} = 4\), we find \(b = 2\), and \(a = \sqrt{11}\). Substituting into the asymptote formula:

• \(y - 3 = \pm \frac{2}{\sqrt{11}}(x - 2)\)

These are the equations for the asymptotes, showing the slopes \(\pm \frac{2}{\sqrt{11}}\). Graphically representing these asymptotes along with the hyperbola provides a complete picture of how the hyperbola is shaped.

Foci

The foci of a hyperbola are two fixed points located inside each "arm" of the hyperbola. These points are crucial as they help define the hyperbola's shape. The line segments connecting each focus to any point on the hyperbola will have a constant difference in lengths. To find the coordinates of the foci for a hyperbola defined by \(\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1\), one needs to calculate \(c\) using the formula:\[c = \sqrt{a^{2} + b^{2}}\] For our equation, \(a^{2} = 11\) and \(b^{2} = 4\), hence:

• \(c^{2} = 11 + 4 = 15\)
• \(c = \sqrt{15}\)

The foci are located along the transverse axis at distances \(c\) from the center. Thus for the center \((2, 3)\), the foci are at:

• \((2 - \sqrt{15}, 3)\)
• \((2 + \sqrt{15}, 3)\)

These points are positioned horizontally because our hyperbola opens left and right. The knowledge of foci helps in understanding the geometric properties and eccentricity of the hyperbola.