Question

write the equation of the ellipse in standard form, and identify the end points of the major and minor axes as well as the foci.
equation in standard form:
end points of major axis:
end points of minor axis:
foci are at:

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse with a vertical major axis is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (\(a>b>0\)), where \((h,k)\) is the center, \(a\) is the semi - major axis length, \(b\) is the semi - minor axis length, and \(c=\sqrt{a^{2}-b^{2}}\) (distance from the center to the foci).
For the given equation \(\frac{(x - 2)^2}{4}+\frac{(y - 1)^2}{16}=1\), we have \(h = 2\), \(k = 1\), \(b^{2}=4\) (so \(b = 2\)), \(a^{2}=16\) (so \(a = 4\)).

Step2: Calculate \(c\)

Using the formula \(c=\sqrt{a^{2}-b^{2}}\), substitute \(a = 4\) and \(b = 2\).
\(c=\sqrt{16 - 4}=\sqrt{12}=2\sqrt{3}\)

Step3: Find the center

The center of the ellipse is \((h,k)=(2,1)\)

Step4: Find the endpoints of the major axis

Since the major axis is vertical (because \(a^{2}\) is under the \(y\) - term), the endpoints of the major axis are \((h,k\pm a)\).
Substitute \(h = 2\), \(k = 1\), \(a = 4\).
The endpoints are \((2,1 + 4)=(2,5)\) and \((2,1-4)=(2,-3)\)

Step5: Find the endpoints of the minor axis

Since the minor axis is horizontal (because \(b^{2}\) is under the \(x\) - term), the endpoints of the minor axis are \((h\pm b,k)\)
Substitute \(h = 2\), \(k = 1\), \(b = 2\)
The endpoints are \((2+2,1)=(4,1)\) and \((2 - 2,1)=(0,1)\)

Step6: Find the foci

Since the major axis is vertical, the foci are \((h,k\pm c)\)
Substitute \(h = 2\), \(k = 1\), \(c = 2\sqrt{3}\)
The foci are \((2,1+2\sqrt{3})\) and \((2,1 - 2\sqrt{3})\)

Answer:

Equation in standard form: \(\frac{(x - 2)^2}{4}+\frac{(y - 1)^2}{16}=1\)
End points of major axis: \((2,5),(2,-3)\)
End points of minor axis: \((4,1),(0,1)\)
Foci are at: \((2,1 + 2\sqrt{3}),(2,1-2\sqrt{3})\)