Question

score on last try: 0 of 1 pts. see details for more.
get a similar question you can retry this question below
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

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

Step2: Analyze equation \(a\): \(\frac{(x - 2)^{2}}{9}+\frac{(y + 1)^{2}}{16}=1\)

Here, \(h = 2\), \(k=-1\), \(a^{2}=9\) (\(a = 3\)), \(b^{2}=16\) (\(b = 4\)). Since \(b>a\), the major axis is vertical. The center is \((2,-1)\).

Step3: Analyze equation \(b\): \(\frac{(x - 2)^{2}}{16}+\frac{(y - 1)^{2}}{9}=1\)

Here, \(h = 2\), \(k = 1\), \(a^{2}=16\) (\(a=4\)), \(b^{2}=9\) (\(b = 3\)). Since \(a>b\), the major axis is horizontal. The center is \((2,1)\).

Step4: Analyze equation \(c\): \(\frac{(x + 2)^{2}}{9}+\frac{(y - 1)^{2}}{16}=1\)

Here, \(h=-2\), \(k = 1\), \(a^{2}=9\) (\(a = 3\)), \(b^{2}=16\) (\(b = 4\)). Since \(b>a\), the major axis is vertical. The center is \((-2,1)\).

Step5: Analyze equation \(d\): \(\frac{(x + 2)^{2}}{16}+\frac{(y + 1)^{2}}{9}=1\)

Here, \(h=-2\), \(k=-1\), \(a^{2}=16\) (\(a = 4\)), \(b^{2}=9\) (\(b = 3\)). Since \(a>b\), the major axis is horizontal. The center is \((-2,-1)\).

Step6: Match the centers and axes

• For graph \(a\) (center \((-2,1)\) with vertical major - axis), it matches equation \(c\).
• For graph \(b\) (center \((-2,-1)\) with horizontal major - axis), it matches equation \(d\).

Answer:

Graph \(a\) matches equation \(c\), Graph \(b\) matches equation \(d\)