find the vertices, foci, equations of the asymptotes, length of LR, and the graph of the hyperbola x^2-4y^2+4x+24y=28
in standard form, you have
x^2+4x - 4(y^2-6y) = 28
x^2+4x+4 - 4(y^2-6y+9) = 28+4-4*9
(x+2)^2 - 4(y-3)^2 = -4
(y-3)^2 - (x+2)^2/4 = 1
Now use that with your standard properties of hyperbolas.
check with this:
http://www.wolframalpha.com/input/?i=hyperbola+(y-3)%5E2+-+(x%2B2)%5E2%2F4+%3D+1
To find the vertices, foci, equations of the asymptotes, and the length of the Latus Rectum (LR) of a hyperbola, we need to rewrite the given equation in standard form:
x^2 - 4y^2 + 4x + 24y = 28
First, we'll complete the square for both the x and y terms separately. Let's start with the x terms:
x^2 + 4x = 4(y^2 - 6y) + 28
To complete the square for x, we add (4/2)^2 = 4 to both sides:
x^2 + 4x + 4 = 4(y^2 - 6y) + 28 + 4
Simplifying further:
(x + 2)^2 = 4(y^2 - 6y + 9) + 32
(x + 2)^2 = 4(y - 3)^2 + 32
Next, we'll rewrite this equation in standard form by dividing both sides by 4:
(x + 2)^2 / 4 - (y - 3)^2 / 8 = 1
Now, we can identify the key characteristics of the hyperbola:
1. Vertices: The vertices are given by the terms (h, k ± a), where (h, k) is the center of the hyperbola and "a" is the distance from the center to either vertex.
From the equation, we can see that (h, k) = (-2, 3). Looking at the equation, we can determine that a^2 = 4, so a = 2. Therefore, the vertices are (-2, 3 ± 2), which gives us the vertices (-2, 1) and (-2, 5).
2. Foci: The foci are given by the terms (h, k ± c), where c^2 = a^2 + b^2, and a and b are the lengths of the semi-major and semi-minor axes.
From the equation, we have a = 2. To find b, we can rearrange the equation to b^2 = a^2(e^2 - 1) and substitute the values:
b^2 = 2^2(1 + 8) = 36
b = 6
Now, we can find c:
c^2 = a^2 + b^2
c^2 = 4 + 36
c^2 = 40
c ≈ √40 ≈ 6.32
Therefore, the foci are (-2, 3 ± 6.32), which gives us the foci (-2, -3.32) and (-2, 9.32).
3. Equations of Asymptotes: The equations of the asymptotes are given by y = ± (b/a)(x - h) + k.
Substituting the values, we have:
y = ± (6/2)(x + 2) + 3
Simplifying further, we get:
y = ± 3(x + 2) + 3
which gives us the asymptotes y = 3x + 9 and y = -3x - 3.
4. Length of Latus Rectum (LR): The length of the latus rectum is given by 2b^2/a.
Substituting the values, we have:
2(6)^2 / 2 = 72 / 2 = 36
Therefore, the length of the latus rectum is 36.
5. Graph: The graph of the hyperbola can be plotted using the identified characteristics. It will have a vertical orientation due to y^2 term having a coefficient of -1, and the center will be at (-2, 3). The vertices are at (-2, 1) and (-2, 5), the foci are at (-2, -3.32) and (-2, 9.32), and the asymptotes are y = 3x + 9 and y = -3x - 3.