What are the vertices of the hyperbola x^2/9 - y^2/4=1
I'm not sure of how to start or what equation to use.
Show steps please.
just set y=0. x = ±3
So, the vertices are at (±3,0)
You know that it has to be y=0, because if x=0, -y^2/4 = 1, meaning y^2 has to be negative.
Thank you so much...
To determine the vertices of the hyperbola, we need to rewrite the equation in standard form. The equation of a hyperbola in standard form is:
(x - h)^2/a^2 - (y - k)^2/b^2 = 1 (for a horizontal hyperbola)
or
(y - k)^2/a^2 - (x - h)^2/b^2 = 1 (for a vertical hyperbola)
Comparing your equation, x^2/9 - y^2/4 = 1, with the standard form equations, we can see that the center of the hyperbola is at the origin (h = 0, k = 0) and the hyperbola is horizontal.
To rewrite the given equation in standard form, we can write it as:
x^2/3^2 - y^2/2^2 = 1
Now, we can see that a^2 = 3^2 = 9 and b^2 = 2^2 = 4. Therefore, the values of a and b are √9 = 3 and √4 = 2, respectively.
The vertices of the hyperbola can be found by adding and subtracting "a" from the center. In this case, since the center is at (0, 0), the vertices will be (±a, 0).
Substituting "a" with 3, the vertices will be (±3, 0).
Therefore, the vertices of the hyperbola x^2/9 - y^2/4 = 1 are (-3, 0) and (3, 0).