Write an equation for a hyperbola with foci at (–2, –2) and (8, –2), and asymptote with slope 4/3
(x-3)
To write the equation of a hyperbola given its foci and asymptotes, we can follow these steps:
Step 1: Determine the center of the hyperbola.
The center of the hyperbola is the midpoint between the foci. In this case, the foci are (–2, –2) and (8, –2). We can find the center by averaging the x-coordinates and the y-coordinates of the foci:
Center = ((–2 + 8)/2, (–2 + –2)/2)
= (6/2, –4/2)
= (3, –2)
Step 2: Find the distance from the center to the foci.
The distance from the center to the foci is called the "c" value of the hyperbola. We can calculate it by using the distance formula:
c = √((8 - 3)^2 + (–2 - –2)^2)
= √(5^2 + 0^2)
= √25
= 5
Step 3: Determine the equation of the asymptotes.
The equation of the asymptotes takes the form y = mx + b, where "m" is the slope of the asymptotes. We are given that the asymptotes have a slope of 4/3. Since the asymptotes pass through the center of the hyperbola (3, –2), we can use the point-slope form to find the equation of each asymptote:
Asymptote 1: y - (–2) = (4/3)(x - 3)
⇒ y + 2 = (4/3)(x - 3)
Asymptote 2: y - (–2) = -(4/3)(x - 3)
⇒ y + 2 = -(4/3)(x - 3)
Step 4: Write the equation of the hyperbola.
The standard form equation for a hyperbola with center (h, k), foci (h + c, k) and (h - c, k), and with asymptotes y = mx + b and y = -mx + b, is given by:
(((x - h)^2) / a^2) - (((y - k)^2) / b^2) = 1
In this case, the center is (3, –2), and the distance from the center to the foci is c = 5.
The equation of the hyperbola is:
(((x - 3)^2) / a^2) - (((y + 2)^2) / b^2) = 1
However, to find the values of a and b, we need to relate them to the distance between the center and the foci.
For a hyperbola, we have the relationship:
c^2 = a^2 + b^2
Since c = 5, we can substitute this value into the equation above and solve for a:
(5)^2 = a^2 + b^2
25 = a^2 + b^2
We also know that the slopes of the asymptotes, m, and the ratios a/b are connected through the equation:
m = ±(b / a)
In this case, m = 4/3. So, we can substitute this value and solve for b:
4/3 = ±(b / a)
Simplifying the equation further:
4a = ±3b
Now, we can substitute the expression for b from the equation above into the equation a^2 + b^2 = 25:
a^2 + (4a/3)^2 = 25
a^2 + 16a^2/9 = 25
9a^2 + 16a^2 = 225
25a^2 = 225
a^2 = 225 / 25
a^2 = 9
a = ±3
Substituting the value of a = ±3 back into the equation 4a = ±3b, we have:
4 * 3 = ±3b or 4 * (-3) = ±3b
12 = ±3b or -12 = ±3b
Dividing by 3:
±4 = b or ±(-4) = b
So, we have two sets of possible values for a and b:
(3, 4) and (3, -4)
Therefore, the equation of the hyperbola is:
(((x - 3)^2) / 3^2) - (((y + 2)^2) / 4^2) = 1
Simplified Equation: ((x - 3)^2) / 9 - ((y + 2)^2) / 16 = 1
To write an equation for a hyperbola given its foci and asymptote, we can use the standard form equation:
[(x - h)² / a²] - [(y - k)² / b²] = 1
Where (h, k) represents the center of the hyperbola, a is the distance from the center to a vertex, and b is the distance from the center to a co-vertex.
First, let's find the center of the hyperbola. The center is the midpoint between the two foci. We can find the coordinates of the center by averaging the x-coordinates and the y-coordinates of the foci:
Center x-coordinate = (–2 + 8) / 2 = 6 / 2 = 3
Center y-coordinate = (–2 + –2) / 2 = –4 / 2 = –2
So, the center of the hyperbola is (3, –2).
Next, let's find the distance from the center to a vertex. The distance from the center to a vertex is denoted by a. We can find a using the distance formula:
Distance between foci = 8 - (-2) = 10
a = (Distance between foci) / 2 = 10 / 2 = 5
Now, let's find the distance from the center to a co-vertex. The distance from the center to a co-vertex is denoted by b. We can find b using the relationship between the slope of the asymptote and the ratio of a and b. The slope of the asymptote in this case is 4/3, which means that a/b = 3/4.
a = 5 (from above)
a/b = 3/4
(5/b) = (3/4)
Cross-multiplying, we get:
4a = 3b
4(5) = 3b
20 = 3b
b = 20/3
Therefore, the equation for the hyperbola with foci at (–2, –2) and (8, –2), and asymptote with slope 4/3 is:
[(x - 3)² / 5²] - [(y + 2)² / (20/3)²] = 1
the center is midway between the foci, at (3,-2) and the axis is horizontal, so the equation is
(x-3)^2/a^2 - (y+2)^2/b^2 = 1
So now use the asymptotes to determine a and b.