Question

match the graph to its equation.
$\frac{(x - 9)^2}{16}+\frac{(y + 1)^2}{16}=1$
$\frac{(x + 9)^2}{16}+\frac{(y - 1)^2}{16}=1$
$\frac{(x - 9)^2}{16}+\frac{(y - 1)^2}{16}=1$
$\frac{(x + 9)^2}{16}+\frac{(y + 1)^2}{16}=1$

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\), where \((h,k)\) is the center of the ellipse.

Step2: Analyze each equation for the center

• For \(\frac{(x - 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\), the center is \((9,-1)\).
• For \(\frac{(x + 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\), the center is \((-9,1)\).
• For \(\frac{(x - 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\), the center is \((9,1)\).
• For \(\frac{(x + 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\), the center is \((-9,-1)\).

Step3: Match the center with the graph

Assuming the standard coordinate - grid, if we consider the position of the center of the ellipse (since \(a = b = 4\), it's a circle in this case as \(a=b\) for the given equations).
Let's assume the first graph (top - most) has a center in the first quadrant (if we consider positive \(x\) and negative \(y\) as per the first equation's center \((9,-1)\) is not a perfect match. Wait, no, if we consider the general position:
The equation \(\frac{(x - 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) has center \((9,1)\).
The equation \(\frac{(x + 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) has center \((-9,1)\).
The equation \(\frac{(x - 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) has center \((9,-1)\).
The equation \(\frac{(x + 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) has center \((-9,-1)\).
If we assume the order of the graphs from top - to - bottom as \(A\), \(B\), \(C\), \(D\)

• For the equation \(\frac{(x - 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) (center \((9,1)\)):
• For the equation \(\frac{(x + 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) (center \((-9,1)\)):
• For the equation \(\frac{(x - 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) (center \((9,-1)\)):
• For the equation \(\frac{(x + 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) (center \((-9,-1)\)):

Since the general form of a circle (as \(a = b\)) is \((x - h)^{2}+(y - k)^{2}=r^{2}\) (here \(r = 4\)).
If we assume the first graph (top - most) is for the equation \(\frac{(x - 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) (center \((9,1)\)), the second graph is for \(\frac{(x + 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\) (center \((-9,1)\)), the third graph is for \(\frac{(x - 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) (center \((9,-1)\)) and the bottom - most graph is for \(\frac{(x + 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\) (center \((-9,-1)\))

Answer:

Assuming the order of graphs from top - to - bottom as \(A\), \(B\), \(C\), \(D\)

• \(A\): \(\frac{(x - 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\)
• \(B\): \(\frac{(x + 9)^{2}}{16}+\frac{(y - 1)^{2}}{16}=1\)
• \(C\): \(\frac{(x - 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\)
• \(D\): \(\frac{(x + 9)^{2}}{16}+\frac{(y + 1)^{2}}{16}=1\)