Question
Find the equation of hyperbola whose foci are at(10,2) (0,2)and whose eccentricity is10/3
Answers
Steve
The center of the graph is at (5,2), so the equation will be
(x-5)^2/a^2 - (y-2)^2/b^2 = 1
Now, we have
c = 5
e = c/a = 10/3, so a = 3/2
a^2+b^2=c^2, so
9/4 + b^2 = 100/4, so b^2 = 91/4
The equation is thus
(x-5)^2/(9/4) - (y-2)^2/(91/4) = 1
or
4(x-5)^2/9 - 4(y-2)^2/91 = 1
Confirm this at
http://www.wolframalpha.com/input/?i=hyperbola+%28x-5%29^2%2F%289%2F4%29+-+%28y-2%29^2%2F%2891%2F4%29+%3D+1
(x-5)^2/a^2 - (y-2)^2/b^2 = 1
Now, we have
c = 5
e = c/a = 10/3, so a = 3/2
a^2+b^2=c^2, so
9/4 + b^2 = 100/4, so b^2 = 91/4
The equation is thus
(x-5)^2/(9/4) - (y-2)^2/(91/4) = 1
or
4(x-5)^2/9 - 4(y-2)^2/91 = 1
Confirm this at
http://www.wolframalpha.com/input/?i=hyperbola+%28x-5%29^2%2F%289%2F4%29+-+%28y-2%29^2%2F%2891%2F4%29+%3D+1