Let equation of an hyperbola be y^2-4x^2+4y+24x-41=0
a. Find the standard form
b. Find the center
c. Find the vertices
d. Find the foci
e. Find the slant asymptotes
To find the answers to all these questions, we need to manipulate the given equation and observe its properties. Let's go through each question step by step.
a. Find the standard form:
The standard form of a hyperbola equation is given by ((y-k)^2)/b^2 - ((x-h)^2)/a^2 = 1, where (h, k) is the center, and a and b are the distances from the center to the vertices along the x and y-axes, respectively.
In this case, we need to complete the square for both the x and y terms. Rearranging and grouping the x and y terms, we get:
(y^2 + 4y) - (4x^2 + 24x) = 41
Now, complete the square for each term separately. For the y-term, we add (4/2)^2 = 4 to complete the square, and for the x-term, we add (24/2)^2 = 144 to complete the square. This gives us:
(y^2 + 4y + 4) - (4x^2 + 24x + 144) = 41 + 4 - 144
(y + 2)^2 - 4(x + 6)^2 = -99
The equation is now in standard form.
b. Find the center:
From the standard form, we can identify the center as (-h, -k). In this case, comparing the equation with the standard form, we have h = -6 and k = -2. Therefore, the center of the hyperbola is at (-(-6), -(-2)), which simplifies to (6, 2).
c. Find the vertices:
Since the vertices lie on the transverse axis, which is the line passing through the center, we need to find the values of a and b.
From the standard form, we can see that a^2 = 1 and b^2 = -99. However, since a is the distance from the center to the vertices along the x-axis, and in this case a^2 = 1, we can conclude that a = 1.
Now, to find the vertices, we add or subtract a to/from the x-coordinate of the center. This gives us two vertices:
Vertex 1: (6 + 1, 2) = (7, 2)
Vertex 2: (6 - 1, 2) = (5, 2)
d. Find the foci:
The distance from the center to the foci is denoted by c. To find c, we use the relation c^2 = a^2 + b^2, where a = 1 and b^2 = -99 from the standard form.
c^2 = 1 + (-99)
c^2 = -98
Since the foci are also located on the transverse axis, their coordinates will be (x ± c, y), where (x, y) is the center. Plugging in the values, we get:
Foci 1: (6 + √(-98), 2) = (6 + i√98, 2)
Foci 2: (6 - √(-98), 2) = (6 - i√98, 2)
Note: Since b^2 is negative, the foci will have imaginary coordinates.
e. Find the slant asymptotes:
To find the slant asymptotes, we can use the graphing of the hyperbola. Slopes of asymptotes can be approximated by the ratio b/a. In this case, b = √(-99) = i√99 and a = 1.
The slopes of the asymptotes will be ±(b/a) = ±(i√99/1) = ±i√99.
Therefore, the slant asymptotes are y = ±i√99x + 2.