Question

help please

Identify the equation of the hyperbola with its center at (3, -1), its vertex at (5, -1), and an asymptote of 2y = 3x – 11.

9(x - 3)2 - 4(y + 1)2 -36 = 0
4(y +1)2 - 9(x -3)2 = 36
(y + 3)2 + (x -1)2 + 36 = 0
4(y - 2)2 - 9(y + 1)2 = 36

Answer 1

Center is (3,-1) We need an x-3 and a y+1. This immediately rules out hte bottom two.

Now what?

Answer 2

center at (3, -1) and vertex at (5,-1)
The center and the vertex have the same y value which implies these two points lie on a horizontal line (y = -1). Therefore, the transverse axis is horizontal with the two branches of the hyperbola opening sideways. The general equation of a hyperbola when the transverse axis is horizontal is:
(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1 ----- (1)
where (h,k) is the center of the hyperbola;
a is the length of the semi-transverse axis (which is the distance between the center and the vertex);
b is the length of the semi-conjugate axis.

Here h = 3, y = -1, a = distance between the center and the vertex = 5 - 3 = 2. Substitute in (1):
(x-3)^2 / 4 - (y+1)^2 / b = 1 ---- (2)

When the transverse axis is horizontal, the equation of the asymptotes are:
y = +/- b/a * (x-h) + k
plug in the numbers (choose just the + asymptote):
y = b/2 * (x-3) - 1 ----- (3)
The given asymptote is: 2y = 3x - 11 or
y = 3x/2 - 11/2 = 3/2 * (x-3) + 9/2 - 11/2.
y = 3/2 * (x-3) - 1 ---- (4)
comparing (3) and (4):
b = 3. Put this in (2):
(x-3)^2 / 4 - (y+1)^2 / 9 = 1
Multiply throughout by 36
9(x-3)^2 - 4(y+1)^2 = 36
9(x-3)^2 - 4(y+1)^2 - 36 = 0

Answer 3

Wow! Also, for fast test takers...

2y = 3x – 11.

\(2\ne3\), so that rules out the third one. It must be #4.

Answer 4

With the explanation, the answer seems longer but the math part can be done fairly quickly.

Answer 5

No objection to the arithmetic. Some students, particularly those preparing for GMAT or LSAT, for example, need to be fast more than anything else. Good to see that both are possible. :-)