Find the standard form of the equation of the hyperbola with vertices (0±3) and asymptotes (y=±3/2x)
Why did the hyperbola go to therapy? Because it couldn't find its "center" and had trouble "focusing"!
Now, let's get back to business. The standard form of the equation of a hyperbola with vertices (0±3) and asymptotes (y=±3/2x) is:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
First, let's find the value of h and k, which represent the center of the hyperbola. In this case, the x-coordinate of the center is 0 since it is given as (0±3). The y-coordinate is also 0 since the asymptotes' equations (y=±3/2x) pass through the origin.
So, the center of the hyperbola is (h, k) = (0, 0).
Now, let's find the values of a and b, which represent the distances from the center to the vertices. In this case, the distance from the center to the vertices is 3, so a = 3.
We also know that the asymptotes have a slope of ±3/2. For a hyperbola, the slope of the asymptotes is given by b/a. So, we can solve for b:
b/a = 3/2
Cross-multiplying, we get:
2b = 3a
2b = 3(3)
2b = 9
b = 9/2
Now we can plug these values into the standard form equation:
(x-0)^2/3^2 - (y-0)^2/(9/2)^2 = 1
Simplifying, we have:
x^2/9 - y^2/(81/4) = 1
To eliminate the denominator, we can multiply both sides by 9 to get:
4x^2 - 9y^2/81 = 9
Finally, we can simplify further:
4x^2 - y^2/9 = 1
So, in standard form, the equation of the hyperbola is:
4x^2 - y^2/9 = 1
Now that we've solved this question, let's take a well-deserved break and enjoy some "asymptote-ically" funny jokes!
To find the standard form of the equation of a hyperbola, we need to gather the necessary information. In this case, we are given the vertices and the asymptotes.
The standard form of the equation of a hyperbola with horizontal transverse axis is:
(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1,
where (h, k) represents the center of the hyperbola, and a and b are the distances from the center to the vertices.
1. Determine the center of the hyperbola: The center is the midpoint between the vertices. Since the vertices are (0±3), the center is at (0, 0).
2. Find the value of a: The distance from the center to each vertex is given as 3. So, a = 3.
3. Find the value of b: The distance from the center to the asymptote is given as 3/2. Since the asymptotes are lines passing through the center, it means that b = 3/2.
Now, we can substitute these values into the equation:
(x-0)^2 / 3^2 - (y-0)^2 / (3/2)^2 = 1.
Simplifying this equation gives us the standard form of the equation:
x^2 / 9 - y^2 / (9/4) = 1.
Therefore, the standard form of the equation of the hyperbola with vertices (0±3) and asymptotes (y=±3/2x) is x^2/9 - y^2/(9/4) = 1.
To find the standard form equation of a hyperbola, we need to use the given information, which includes the vertices and asymptotes.
1. The standard form of a hyperbola with horizontal transverse axis is:
((x-h)^2)/(a^2) - ((y-k)^2)/(b^2) = 1
Where the center of the hyperbola is (h,k), the semi-major axis is a, and the semi-minor axis is b.
2. From the given information, we know that the vertices of the hyperbola are (0±3), which means the center of the hyperbola is at the origin (0, 0).
3. The distance between the center and each vertex represents the value of a. In this case, a = 3.
4. The equation of the asymptotes is in the form y = ±(b/a)x, where b represents the value of the semi-minor axis. In this case, the asymptotes are y = ±(3/2)x, which means b/a = 3/2.
5. To find the value of b, we can solve the equation b/a = 3/2.
b/a = 3/2
b/3 = 3/2 (substituting a = 3)
b = 9/2
6. Now we have all the necessary values to write the equation of the hyperbola:
((x-0)^2)/(3^2) - ((y-0)^2)/(9/2)^2 = 1
x^2/9 - y^2/(81/4) = 1
To get rid of the fraction, we can multiply through by 9:
4x^2 - 9y^2/81 = 9
4x^2 - 9y^2/9 = 81
Simplifying further:
4x^2 - y^2/9 = 9
This is the standard form of the equation of the hyperbola.
form y = (3/2)x
b:a = 3/2, but a = 3
b/3 = 3/2
2b = 9
b = 9/2
hyperbola:
x^2 /9 - y^2/(81/4) = 1
or
x^2/9 - 4y^2/81 = 1
or
9x^2 - 4y^2 = 81
whichever form your text considers standard form