SOLUTION: Find an equation in standard form for the hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
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The work I have:
Since c is the distance between the f
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-> SOLUTION: Find an equation in standard form for the hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
.
The work I have:
Since c is the distance between the f
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Question 860695: Find an equation in standard form for the hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
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The work I have:
Since c is the distance between the foci and center, c=7-1=6
c^2=a^2+b^2=36
Since e=c/a=2, c/a=2/1=6/3
Since c/a= sqrt.(a^2-b^2)/a, 6/3= sqrt.(9-b^2)/3.
36=9-b^2
27=-b^2
-27=b^2
So the equation would be [(x-1)^2]/9-[(y+4)^2]/-27.
but the -27 would change the sign of the y to +
so it'd be [(x-1)^2]/9+[(y+4)^2]/-27
But isn't the standard form for hyperbolas [(x-h)^2]/a^2-[(y+k)^2]/b^2?
How can it be a + and still be a hyperbola?
Please tell me if I'm doing something wrong. Answer by ewatrrr(24785) (Show Source):
Hi,
re TY: Foci units units up and down from center
there is NO minus to consider (You may be thinking of the ellipse format)
hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
Center, Vertices and foci along y = -4
Yes!
2 = c/a and Yes, c = 6 and a = 3
sqrt(9+ 27) = 6, b^2 = 27 |Foci units units up and down from center
