|
Question 422433: Choose the equation that best represents a hyperbola with vertices (0, –1) and (4, –1), and foci (–1, –1) and (5, –1).
(a) (x-2)^2/4 - (y+1)^2/5 = 1
(b) (x-2)^2/25 - (y+1)^2/16 =1
(c) (x+2)^2/4 + (y+1)^2/5 = 1
(d) x^2/4 - y^2/5 = 1
Found 2 solutions by Edwin McCravy, ewatrrr:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! (a) (x-2)^2/4 - (y+1)^2/5 = 1
(b) (x-2)^2/25 - (y+1)^2/16 =1
(c) (x+2)^2/4 + (y+1)^2/5 = 1
(d) x^2/4 - y^2/5 = 1
We can eliminate (c) immediately because the + sign makes it the
equation of an ellipse, not a hyperbola.
The vertices are points on the curve, so we substitute the vertices
(0, –1) and (4, –1) to see if (a) could be the correct answer:
(x-2)^2/4 - (y+1)^2/5 = 1
(0-2)^2/4 - (-1+1)^2/5 = 1
4/4 - 0 = 1
1 = 1
(x-2)^2/4 - (y+1)^2/5 = 1
(4-2)^2/4 - (-1+1)^2/5 = 1
4/4 - 0 = 1
1 = 1
So (a) could be the answer, but we aren't sure.
so we substitute the vertices
(0, –1) and (4, –1) to see if (b) could be the correct answer:
(x-2)^2/25 - (y+1)^2/16 = 1
(0-2)^2/25 - (-1+1)^2/16 = 1
4/25 - 0 = 1
4/25 = 1
So we have ruled out (b) without even bothering to substitute
the other vertex.
The only other choice is (d), so we substitute the vertices
(0, –1) and (4, –1) to see if (d) could be the correct answer:
x^2/4 - y^2/5 = 1
0^2/4 - (-1)/5 = 1
0 + 1/5 = 1
1/5 = 1
So we have ruled out (d) without even bothering to substitute
the other vertex.
So the only possible answer is (a).
Edwin
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi
hyperbola with vertices (0, –1) and (4, –1), and foci (–1, –1) and (5, –1).
Note: Line of symmetry is y = -1, Hyperbola opens right and left
Standard Form of an Equation of an Hyperbola is 
where Pt(h,k) is a center with vertices 'a' units right and left of center.
vertices (0,-1) and (4,-1) tell us the Center is (2,-1)
thus: 
foci are 3 units from center: "sqrt(4+5)" = 3, b =5
Always recommend sketching it out:
|
|
|
| |
