Answer to Question #238685 in Analytic Geometry for joyce

Question

Find the equation of the hyperbola with foci at (1,2) (1,6) and vertices at (1,3) (1,5)

Answer 1

Foci $(h, k\pm c),$

Vertices $(h, k\pm a)$

Equation

$$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$$

$$a^2+b^2=c^2$$

Foci $(1,2),(1,6)$

Vertices $(1,3),(1,5)$

Substitute

$$h=1, k-c=2, k+c=6$$

$$h=1, k-a=3, k+a=5$$

Then $h=1, k=4, c=2, a=1$

$$b^2=c^2-a^2=(2)^2-(1)^2=3$$

The equation of the hyperbola is

$$\dfrac{(y-4)^2}{1}-\dfrac{(x-1)^2}{3}=1$$