Problem 57 - Algebra and Trigonometry

Question

In Exercises \(51-60,\) convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$ 49 x^{2}+16 y^{2}+98 x-64 y-671=0 $$

Step-by-Step Solution

Answer

The given equation is equivalent to \((\frac {(x+1)^{2}}{4}) + (\frac {(y-2)^{2}}{49}) = 1\). The foci are located at (-1, 2+ \sqrt{45}) and (-1, 2- \sqrt{45}).

Step 1: Rearrange the Equation

Rearrange the terms of the given equation to group \(x\) terms and \(y\) terms together: \(49x^{2}+98x+16y^{2}-64y=671.\)

Step 2: Complete the Square for Both Variables

To convert to the standard form of the ellipse, complete the square for both variables \(x\) and \(y\). This involves adding and subtracting the square of half the coefficient of \(x\) and \(y\). The equation will now look like this: \((49x^{2}+98x+ 49)+ (16y^{2}-64y+64) - 49 - 64 = 671\). This simplifies to: \((7x+7)^{2}+(4y-8)^{2}-113 = 671.\)

Step 3: Convert the Equation to Standard Form

Get the standard form of the ellipse by making the right side of the equation equal to 1. Simplify \((7x+7)^{2}+(4y-8)^{2} = 784\), then divide through by 784 to get: \(\[(\frac {(x+1)^{2}}{4}) + (\frac {(y-2)^{2}}{49}) = 1\].

Step 4: Identifying the Foci

The foci of the ellipse can be located using the equation \(c = \sqrt{a^{2} - b^{2}}\), where \(a\) is the semi-major axis (sqrt of the larger denominator), \(b\) is the semi-minor axis (sqrt of the smaller denominator), and \(c\) is the distance from the center to a single focus. Here, \(a = 7\), \(b = 2\), so \(c = \sqrt{45}\). The foci are spaced \(c\) units above and below the center of the ellipse, \(c\) being a vertical distance because this is a vertically oriented ellipse. So the two foci are located at \((h, k \pm c)\), or \((-1,2 \pm \sqrt{45})\).

Step 5: Graphing the Ellipse

The graph of this equation forms an ellipse with center at point (-1,2), major axis of 14 along the y-axis and minor axis of 4 along the x-axis. The foci are located at points (-1, 2+ \sqrt{45}) and (-1, 2- \sqrt{45}).