Question

consider the following equation of an ellipse. 36x² + y² + 360x + 10y + 889 = 0 step 3 of 4: find the endpoints of the major and minor axes of this ellipse.

Explanation:

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

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

$$
LATEXBLOCK0
$$

For the \(x\) - part: \((x + 5)^{2}=x^{2}+10x + 25\), so \(36(x^{2}+10x)=36[(x + 5)^{2}-25]=36(x + 5)^{2}-900\)
For the \(y\) - part: \((y + 5)^{2}=y^{2}+10y+25\), so \(y^{2}+10y=(y + 5)^{2}-25\)

The equation becomes:

$$
LATEXBLOCK1
$$

The standard form of an ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\)), where \((h,k)=(-5,-5)\), \(a = 6\), \(b = 1\)

Step2: Find endpoints of major and minor axes

For the major axis (parallel to \(y\) - axis since \(a>b\)):
The endpoints are \((h,k\pm a)\)
Substitute \(h=-5\), \(k=-5\), \(a = 6\)
\((-5,-5 + 6)=(-5,1)\) and \((-5,-5-6)=(-5,-11)\)

For the minor axis (parallel to \(x\) - axis):
The endpoints are \((h\pm b,k)\)
Substitute \(h=-5\), \(k=-5\), \(b = 1\)
\((-5+1,-5)=(-4,-5)\) and \((-5 - 1,-5)=(-6,-5)\)

Answer:

Endpoints of major axis: \((-5,1)\) and \((-5,-11)\)
Endpoints of minor axis: \((-4,-5)\) and \((-6,-5)\)