Write equations of hyperbolas from graphs - High school Algebra II Exercises

Find the equation of the following hyperbolas.

\left(-4, 1\right) and \left(6{,}1\right) are the focuses of the hyperbola.

\dfrac{\left(x-1\right)^2}{9}-\dfrac{\left(y-1\right)^2}{9}=1

\dfrac{\left(x-1\right)^2}{16}-\dfrac{\left(y-1\right)^2}{9}=1

\dfrac{\left(y-1\right)^2}{9}-\dfrac{\left(x-1\right)^2}{16}=1

\dfrac{\left(y-1\right)^2}{9}-\dfrac{\left(x-1\right)^2}{25}=1

\left(-6{,}2\right) and \left(4{,}2\right) are the focuses of the hyperbola.

\dfrac{\left(x-2\right)^2}{25}-\dfrac{\left(y-1\right)^2}{9}=1

\dfrac{\left(x+1\right)^2}{9}-\dfrac{\left(y-2\right)^2}{16}=1

\dfrac{\left(y+1\right)^2}{9}-\dfrac{\left(x-1\right)^2}{16}=1

\dfrac{\left(y-1\right)^2}{9}-\dfrac{\left(x+3\right)^2}{25}=1

\left(-7{,}0\right) and \left(13{,}0\right) are the focuses of the hyperbola.

\dfrac{\left(x-3\right)^2}{9}-\dfrac{\left(y-2\right)^2}{25}=1

\dfrac{\left(x-3\right)^2}{36}-\dfrac{y^2}{64}=1

\dfrac{\left(y-1\right)^2}{36}-\dfrac{\left(x+1\right)^2}{64}=1

\dfrac{\left(y-1\right)^2}{9}-\dfrac{\left(x-1\right)^2}{25}=1

\left(-13{,}0\right) and \left(13{,}0\right) are the focuses of the hyperbola.

\dfrac{x^2}{9}-\dfrac{y^2}{144}=1

\dfrac{x^2}{25}-\dfrac{y^2}{144}=1

\dfrac{\left(y-3\right)^2}{9}-\dfrac{\left(x+4\right)^2}{16}=1

\dfrac{\left(y+1\right)^2}{9}-\dfrac{\left(x-1\right)^2}{25}=1

\left(-3, 1\right) and \left(-3{,}11\right) are the focuses of the hyperbola.

\dfrac{\left(x-1\right)^2}{9}-\dfrac{\left(y-1\right)^2}{9}=1

\dfrac{\left(y-6\right)^2}{16}-\dfrac{\left(x+3\right)^2}{9}=1

\dfrac{\left(y+1\right)^2}{144}-\dfrac{\left(x-1\right)^2}{16}=1

\dfrac{\left(y-2\right)^2}{25}-\dfrac{\left(x+1\right)^2}{169}=1

\left(4, -10\right) and \left(4, 10\right) are the focuses of the hyperbola.

\dfrac{\left(x-5\right)^2}{36}-\dfrac{\left(y-1\right)^2}{64}=1

\dfrac{y^2}{36}-\dfrac{\left(x-4\right)^2}{64}=1

\dfrac{\left(y+4\right)^2}{9}-\dfrac{\left(x-3\right)^2}{16}=1

\dfrac{y^2}{36}-\dfrac{\left(x-4\right)^2}{25}=1

\left(-2, 1-\sqrt{2}\right) and \left(-2, 1+\sqrt{2}\right) are the focuses of the hyperbola.

\dfrac{\left(x+7\right)^2}{3}-\dfrac{\left(y-2\right)^2}{25}=1

\left(y-1\right)^2-\left(x+2\right)^2=1

\dfrac{\left(y+7\right)^2}{9}-\dfrac{\left(x-1\right)^2}{16}=1

\left(y+1\right)^2-\left(x-3\right)^2=1