Question 422433
(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
<pre><font face = "consolas" color = "indigo" size = 4><b>
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</pre>