Find the graph of the ellipses that have the following equations.
\dfrac{\left(x-2\right)^2}{4}+\dfrac{\left(y+1\right)^2}{9}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=2, \, b=3,\, h=2, \,k=-1
- The center of the ellipse is \left(2,-1\right)
- The horizontal axis of the ellipse is a segment with a length of 4 from the point \left(0,-1\right) to the point \left(4,-1\right).
- The vertical axis of the ellipse is a segment with a length of 6 from the point \left(2,-4\right) to the point \left(2{,}2\right).
- Since a \lt b, the vertices of the ellipse are \left(2,-4\right) and \left(2{,}2\right).
The graph of the ellipse is as follows:

\dfrac{\left(x-1\right)^2}{4}+\dfrac{\left(y-1\right)^2}{16}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=2, \, b=4,\, h=1, \,k=1
- The center of the ellipse is \left(1{,}1\right)
- The horizontal axis of the ellipse is a segment with a length of 4 from the point \left(-1{,}1\right) to the point \left(3{,}1\right).
- The vertical axis of the ellipse is a segment with a length of 8 from the point \left(1,-3\right) to the point \left(1{,}5\right).
- Since a \lt b, the vertices of the ellipse are \left(1,-3\right) and \left(1{,}5\right).
The graph of the ellipse is as follows:

\dfrac{\left(x+2\right)^2}{1}+\dfrac{y^2}{4}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=1, \, b=2,\, h=-2, \,k=0
- The center of the ellipse is \left(-2{,}0\right)
- The horizontal axis of the ellipse is a segment with a length of 2 from the point \left(-3{,}0\right) to the point \left(-1{,}0\right).
- The vertical axis of the ellipse is a segment with a length of 4 from the point \left(-2,-2\right) to the point \left(-2{,}2\right).
- Since a \lt b, the vertices of the ellipse are \left(-2,-2\right) and \left(-2{,}2\right).
The graph of the ellipse is as follows:

\dfrac{x^2}{9}+\dfrac{\left(y+1\right)^2}{25}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=3, \, b=5,\, h=0, \,k=1
- The center of the ellipse is \left(0{,}1\right)
- The horizontal axis of the ellipse is a segment with a length of 6 from the point \left(-3{,}1\right) to the point \left(3{,}1\right).
- The vertical axis of the ellipse is a segment with a length of 10 from the point \left(0,-4\right) to the point \left(0{,}6\right).
- Since a \lt b, the vertices of the ellipse are \left(0,-4\right) and \left(0{,}6\right).
The graph of the ellipse is as follows:

\dfrac{\left(x+3\right)^2}{16}+\dfrac{\left(y+1\right)^2}{4}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=4, \, b=2,\, h=-3, \,k=-1
- The center of the ellipse is \left(-3,-1\right)
- The horizontal axis of the ellipse is a segment with a length of 8 from the point \left(-7,-1\right) to the point \left(1,-1\right).
- The vertical axis of the ellipse is a segment with a length of 4 from the point \left(-3,-3\right) to the point \left(-3{,}1\right).
- Since a \gt b, the vertices of the ellipse are \left(-7,-1\right) and \left(1,-1\right).
The graph of the ellipse is as follows:

\dfrac{\left(x-1\right)^2}{25}+\dfrac{\left(y+3\right)^2}{9}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question, we have:
- a=5, \, b=3,\, h=1, \,k=3
- The center of the ellipse is \left(1{,}3\right)
- The horizontal axis of the ellipse is a segment with a length of 10 from the point \left(-4{,}3\right) to the point \left(6{,}3\right)..
- The vertical axis of the ellipse is a segment with a length of 6 from the point \left(1{,}0\right) to the point \left(1{,}6\right).
- Since a \gt b, the vertices of the ellipse are \left(-4{,}3\right) and \left(6{,}3\right).
The graph of the ellipse is as follows:

\dfrac{\left(x+2\right)^2}{36}+\dfrac{\left(y-2\right)^2}{9}=1
Given the standard formula of an ellipse:
\dfrac{\left(x-h\right)^2}{a^2}+ \dfrac{\left(y-k\right)^2}{b^2}=1
- The center of the ellipse is \left(h,k\right)
- The horizontal axis of the ellipse is a segment with a length 2a of from the point \left(h-a,k\right) to the point \left(h+a,k\right).
- The vertical axis of the ellipse is a segment with a length of 2b from the point \left(h,k-b\right) to the point \left(h,k+b\right).
- If a \gt b, then the vertices of the ellipse are \left(h\pm a,k\right).
- If a \lt b, then the vertices of the ellipse are \left(h,k \pm b\right).
In this question we have:
- a=6, \, b=3,\, h=-2, \,k=2
- The center of the ellipse is \left(-2{,}2\right)
- The horizontal axis of the ellipse is a segment with a length of 12 from the point \left(-8{,}2\right) to the point \left(4{,}2\right).
- The vertical axis of the ellipse is a segment with a length of 6 from the point \left(-2,-1\right) to the point \left(-2{,}5\right).
- Since a \gt b, the vertices of the ellipse are \left(-8{,}2\right) and \left(4{,}2\right).
The graph of the ellipse is as follows:
