Question

consider the following equation of an ellipse.
36x² + y² + 360x + 10y + 889 = 0
step 4 of 4: find the coordinates of the two foci of this ellipse. round your answer to two decimal places.

Explanation:

Step1: Complete the square for \(x\) and \(y\) terms

Group \(x\) - terms and \(y\) - terms:

$$
LATEXBLOCK0
$$

For the \(x\) - part: \(x^{2}+10x=(x + 5)^{2}-25\). For the \(y\) - part: \(y^{2}+10y=(y + 5)^{2}-25\)

$$
LATEXBLOCK1
$$

Divide by \(36\) to get the standard form \(\frac{(x + 5)^{2}}{1}+\frac{(y + 5)^{2}}{36}=1\)

Step2: Identify \(a\), \(b\) and \(c\)

For the ellipse \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}} = 1\) (\(a>b>0\)), here \(h=-5\), \(k = - 5\), \(a^{2}=36\), \(b^{2}=1\)
Use the formula \(c=\sqrt{a^{2}-b^{2}}\), so \(c=\sqrt{36 - 1}=\sqrt{35}\approx5.92\)

Step3: Find the foci coordinates

Since the major axis is parallel to the \(y\) - axis (because \(a^{2}\) is under the \(y\) - term), the foci are \((h,k\pm c)\)
Substitute \(h=-5\), \(k=-5\) and \(c\approx5.92\)
The foci are \((-5,-5 - 5.92)\) and \((-5,-5+5.92)\)

Answer:

\((-5,-10.92)\) and \((-5,0.92)\)