Find the equation of the following hyperbolas.
\left(-4, 1\right) and \left(6{,}1\right) are the focuses of the hyperbola.
This is the graph of a horizontal hyperbola. The general formula of a horizontal hyperbola is:
\dfrac{\left(x-h\right)^2}{a^2}-\dfrac{\left(y-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(1{,}1\right) and a=3.
Also:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=5.
Hence:
b^2= 5^2 - 3^2 = 16
Therefore, the standard form of the hyperbola is as below:
\dfrac{\left(x-1\right)^2}{9}-\dfrac{\left(y-1\right)^2}{16}=1
\left(-6{,}2\right) and \left(4{,}2\right) are the focuses of the hyperbola.
This is the graph of a horizontal hyperbola. The general formula of a horizontal hyperbola is:
\dfrac{\left(x-h\right)^2}{a^2}-\dfrac{\left(y-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(-1{,}2\right) and a=3.
Also:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=5.
Hence:
b^2= 5^2 - 3^2 = 16
Therefore, the standard form of the hyperbola is as below:
\dfrac{\left(x+1\right)^2}{9}-\dfrac{\left(y-2\right)^2}{16}=1
\left(-7{,}0\right) and \left(13{,}0\right) are the focuses of the hyperbola.
This is the graph of a horizontal hyperbola. The general formula of a horizontal hyperbola is:
\dfrac{\left(x-h\right)^2}{a^2}-\dfrac{\left(y-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(3{,}0\right) and a=6.
Also:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=10.
Hence:
b^2= 10^2 - 6^2 = 64
Therefore, the standard form of the hyperbola is as below:
\dfrac{\left(x-3\right)^2}{36}-\dfrac{y^2}{64}=1
\left(-13{,}0\right) and \left(13{,}0\right) are the focuses of the hyperbola.
This is the graph of a horizontal hyperbola. The general formula of a horizontal hyperbola is:
\dfrac{\left(x-h\right)^2}{a^2}-\dfrac{\left(y-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(0{,}0\right) and a=5.
Also:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=13.
Hence:
b^2= 13^2 - 5^2 = 144
Therefore, the standard form of the hyperbola is as below:
\dfrac{x^2}{25}-\dfrac{y^2}{144}=1
\left(-3, 1\right) and \left(-3{,}11\right) are the focuses of the hyperbola.
This is the graph of a vertical hyperbola. The general formula of a vertical hyperbola is:
\dfrac{\left(y-h\right)^2}{a^2}-\dfrac{\left(x-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore \left(h,k\right)=\left(-3{,}6\right) and a=4.
Also:
b^2= c^2-a^2
where c is the distance from the center to either focus. Here, we have c=5.
Hence:
b^2= 5^2 - 4^2 = 9
Therefore, the standard form of the hyperbola is as below:
\dfrac{\left(y-6\right)^2}{16}-\dfrac{\left(x+3\right)^2}{9}=1
\left(4, -10\right) and \left(4, 10\right) are the focuses of the hyperbola.
This is the graph of a vertical hyperbola. The general formula of a vertical hyperbola is:
\dfrac{\left(y-h\right)^2}{a^2}-\dfrac{\left(x-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(4{,}0\right) and a=6.
Also we have:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=10.
Hence:
b^2= 10^2 - 6^2 = 64
Therefore, the standard form of the hyperbola is as below:
\dfrac{y^2}{36}-\dfrac{\left(x-4\right)^2}{64}=1
\left(-2, 1-\sqrt{2}\right) and \left(-2, 1+\sqrt{2}\right) are the focuses of the hyperbola.
This is the graph of a vertical hyperbola. The general formula of a vertical hyperbola is:
\dfrac{\left(y-h\right)^2}{a^2}-\dfrac{\left(x-k\right)^2}{b^2}=1
Where \left(h,k\right) is the center of the hyperbola and a is the distance from the center to either vertex. Therefore, \left(h,k\right)=\left(-2{,}1\right) and a=1.
Also:
b^2= c^2-a^2
Where c is the distance from the center to either focus. Here, we have c=\sqrt{2}.
Hence:
b^2= 2 - 1 = 1
Therefore, the standard form of the hyperbola is as below:
\left(y-1\right)^2-\left(x+2\right)^2=1