Find the vertices, foci, and asymptotes of the hyperbola.
16x2 − 9y2 = 144
To find the vertices, foci, and asymptotes of the hyperbola given by the equation 16x^2 - 9y^2 = 144, we need to complete the square for both the x and y terms.
First, let's divide the equation by 144 to simplify it:
x^2/9 - y^2/16 = 1
Now, let's focus on the x-term:
x^2/9 = 1
To complete the square, we need to move the constant term to the other side:
x^2/9 - 1 = 0
Next, we multiply both sides by 9 to get rid of the fraction:
x^2 - 9 = 0
To complete the square, we add (9/2)^2 = 81/4 to both sides:
x^2 - 9 + 81/4 = 0 + 81/4
Simplifying:
(x - 9/2)^2 = 81/4
Similarly, for the y-term:
-y^2/16 = 1
Again, to complete the square, we move the constant term to the other side:
-y^2/16 - 1 = 0
Multiplying both sides by -16 to remove the fraction:
y^2 + 16 = 0
Adding (16/2)^2 = 64 to both sides to complete the square:
y^2 + 16 + 64 = 0 + 64
Simplifying:
(y + 8)^2 = 64
Now we can see that the equation is in the standard form for a hyperbola, [(x-h)^2/a^2] - [(y-k)^2/b^2] = 1.
Comparing the equations, we can identify the values of h, k, a, and b:
For the x-term:
h = 9/2
a^2 = 81/4
For the y-term:
k = -8
b^2 = 64
From this information, we can determine the center of the hyperbola, which is (h, k):
Center = (9/2, -8)
The semi-major and semi-minor axes can be found by taking the square roots of a^2 and b^2, respectively:
Semi-major axis (a) = √(81/4) = 9/2
Semi-minor axis (b) = √(64) = 8
Now we have all the information needed to determine the vertices, foci, and asymptotes of the hyperbola.
Vertices:
The vertices of the hyperbola are located at (h ± a, k), so the vertices are:
(9/2 ± 9/2, -8) = (9, -8) and (0, -8)
Foci:
The foci of the hyperbola are located at (h ± c, k), where c is the distance from the center to each focus. We can find c using the relationship c^2 = a^2 + b^2:
c^2 = (9/2)^2 + 64
c^2 = 81/4 + 64
c^2 = 329/4
c = √(329/4)
Therefore, the foci are located at (h ± √(329/4), k), giving us the foci:
(9/2 ± √(329/4), -8)
Asymptotes:
The slopes of the asymptotes can be found using the equation b/a, so in this case, -8/(9/2).
The equation of the asymptotes is y = mx + b, where m is the slope of the asymptotes and b is the y-intercept.
Using the slope-intercept form, we can find the y-intercept (b) by plugging in the coordinates of the center (h, k).
y = (-8/(9/2))x + b
-8 = (-8/(9/2))(9/2) + b
-8 = -8 + b
b = 0
So, the equations of the asymptotes are:
y = (-8/(9/2))x
y = (16/9)x
In conclusion, the vertices of the hyperbola are (9, -8) and (0, -8), the foci are (9/2 ± √(329/4), -8), and the asymptotes are y = (-8/(9/2))x and y = (16/9)x.