SOLUTION: 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

Algebra.Com
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:



 

RELATED QUESTIONS


Choose the equation that best represents an ellipse for the given foci and co-vertices.

 (answered by Edwin McCravy)

Y^2 − x^2 /4 = 1.What are all the vertices,asymptotes, foci and... (answered by MathLover1)

find and equation for a hyperbola with vertices at (2,3)  and (2,-1) and foci at (2,6)... (answered by lwsshak3)

The question is:
Determine the vertices, asymptotes, and foci of the hyperbola defined... (answered by Nate)

Choose the equation that best represents the ellipse with center (–5, –2), horizontal... (answered by ewatrrr)

Find the equation of the hyperbola with vertices at (-4,2) 
and (0,2) and foci at (-5,2)  (answered by Edwin McCravy)

Write the equation of the hyperbola with foci at (1, 5) and (7, 5) and with vertices at... (answered by lwsshak3)

Identify the center, vertices, and foci of this hyperbola: 

((x-1)^2 / 4) - ((y-3)^2 /  (answered by ewatrrr)