Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the equation 4(x + 2) 2 + (y+4) 2 4 = 1
.Sketch the graph, and include these points and lines, along with the auxiliary rectangle.
2 answers
Cannot sketch graphs on these posts.
You need to work on writing algebra online
4(x + 2)^2 + (y+4)^2/4 = 1
That's not the standard form, which would be
(x+2)^2 / (1/2)^2 + (y+4)^2/2^2 = 1
The other problem is, that this is the equation of an ellipse.
So, fix your equation and maybe we can get a useful answer
I will just remind you that
(x-h)^2/a^2 - (y-k)^2/b^2 = 1 has
c&^2 = a^2 + b^2
center at (h,k)
vertices at (h±a,k)
asymptotes y-k = ±b/a (x-h)
foci at (h±c,k)
4(x + 2)^2 + (y+4)^2/4 = 1
That's not the standard form, which would be
(x+2)^2 / (1/2)^2 + (y+4)^2/2^2 = 1
The other problem is, that this is the equation of an ellipse.
So, fix your equation and maybe we can get a useful answer
I will just remind you that
(x-h)^2/a^2 - (y-k)^2/b^2 = 1 has
c&^2 = a^2 + b^2
center at (h,k)
vertices at (h±a,k)
asymptotes y-k = ±b/a (x-h)
foci at (h±c,k)
2 answers