Question

give the equation for the ellipse graphed above. question help: video 1 video 2 submit question question 4 match the graphs to their equations. \\( \frac{(x-2)^{2}}{9}+\frac{(y+1)^{2}}{16}=1 \\) \\( \frac{(x-2)^{2}}{16}+\frac{(y-1)^{2}}{9}=1 \\) \\( \frac{(x+2)^{2}}{9}+\frac{(y-1)^{2}}{16}=1 \\) \\( \frac{(x+2)^{2}}{16}+\frac{(y+1)^{2}}{9}=1 \\)

Explanation:

Step1: Recall the standard form of an ellipse equation

The standard form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (if major axis is horizontal) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (if major axis is vertical), where \(a>b\).

Step2: Identify the center of the ellipse

From the graph, the center of the ellipse is at \((- 2,1)\). So \(h=-2\) and \(k = 1\).

Step3: Determine the values of \(a\) and \(b\)

Counting the units from the center to the vertices. The major axis is vertical. If we assume \(a\) is the semi - major axis and \(b\) is the semi - minor axis. Looking at the options, for the equation \(\frac{(x + 2)^2}{9}+\frac{(y - 1)^2}{16}=1\), here \(h=-2,k = 1,a = 4,b = 3\) (since \(a^2=16\Rightarrow a = 4\) and \(b^2=9\Rightarrow b=3\))

Answer:

\(\frac{(x + 2)^2}{9}+\frac{(y - 1)^2}{16}=1\)