Question #238685

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


1
Expert's answer
2021-09-20T14:54:31-0400

Foci (h,k±c),(h, k\pm c),

Vertices (h,k±a)(h, k\pm a)

Equation


(yk)2a2(xh)2b2=1\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1

a2+b2=c2a^2+b^2=c^2

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

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

Substitute


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

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

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


b2=c2a2=(2)2(1)2=3b^2=c^2-a^2=(2)^2-(1)^2=3

The equation of the hyperbola is


(y4)21(x1)23=1\dfrac{(y-4)^2}{1}-\dfrac{(x-1)^2}{3}=1


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