find an hyperbola equation with foci (0, +/-8) and asymptotes; y=+/-1/2x?
Why did the hyperbola go to therapy?
Because its foci had some unresolved issues!
Now, let's find the equation of the hyperbola.
Given the foci at (0, 8) and (0, -8) and the asymptotes y = ±(1/2)x, we can determine the standard form equation using the foci and the eccentricity (e) of the hyperbola.
The distance between the foci and the center of the hyperbola is given by the equation:
c = 8
Here's where the eccentricity comes into play:
e = c/a, where a is the distance from the center to the vertices of the hyperbola.
Since the asymptotes are y = ±(1/2)x, we know that the slope of the asymptotes is equal to ±(b/a), where b is the distance from the center to the intersection of the hyperbola with the asymptotes.
From the asymptotes, we can deduce that b = 1/2.
Now, let's solve for a using the value of c we found earlier:
e = c/a
1 = 8/a
a = 8
Now we have the values of a, b, and c, which can be plugged into the standard form equation of a hyperbola to get the final result:
(x - h)²/a² - (y - k)²/b² = 1
Substituting the values:
(x - 0)²/8² - (y - 0)²/(1/2)² = 1
Simplifying further:
x²/64 - y²/1/4 = 1
Multiplying through by 64 to eliminate the fractions:
64x² - 16y² = 64
And there you go! The equation of the hyperbola is 64x² - 16y² = 64.
To find the equation of a hyperbola with foci (0, +/-8) and asymptotes y = +/- 1/2x, we can follow these steps:
Step 1: Identify the center of the hyperbola.
The center of the hyperbola is given by the midpoint between the two foci. In this case, the foci are (0, 8) and (0, -8), so the center is at (0, 0).
Step 2: Determine the distance between the center and the foci.
The distance between the center and each focus is 8 units since the foci are located at (0, +/-8).
Step 3: Find the equation of the asymptotes.
The equation of the asymptotes is given as y = +/- 1/2x. Since the slopes of the asymptotes are equal to +/- 1/2, the value of "a" (distance from center to vertex) is equal to 2 times the distance from the center to one of the foci. In this case, a = 2 * 8 = 16.
Step 4: Determine the value of "b".
To find "b", we can use the relationship a^2 = b^2 + c^2, where "a" is the distance from the center to the vertex, "b" is the distance from the center to the intersection with the asymptotes, and "c" is the distance from the center to each focus. Since we know "a" = 16 and "c" = 8, we can substitute these values into the equation to solve for "b":
16^2 = b^2 + 8^2
256 = b^2 + 64
b^2 = 256 - 64
b^2 = 192
b = √192 = 8√3
Step 5: Write the equation of the hyperbola.
Now that we have the values of "a" = 16 and "b" = 8√3, we can write the equation of the hyperbola in standard form as:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
In our case, since the center is at (0, 0), the equation becomes:
x^2 / 16^2 - y^2 / (8√3)^2 = 1
x^2 / 256 - y^2 / 192 = 1
Therefore, the equation of the hyperbola with foci (0, +/-8) and asymptotes y = +/- 1/2x is x^2 / 256 - y^2 / 192 = 1.
To find the equation of a hyperbola given the 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, since the foci are located at (0, +/-8), the center will be at (0, 0).
Step 2: Find the value of "a."
The distance from the center to each focus is known as "c." In this case, c = 8. The distance from the center to the vertex (end point of the transverse axis) is known as "a." Since the asymptotes have slopes of +/-1/2, the value of "a" can be found using the equation a = |1/(2 * slope of asymptotes)| = |1/(2 * 1/2)| = |1| = 1.
Step 3: Determine the equation.
The standard form of the equation for a hyperbola with the center at (h, k) is given by:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
In this case, our center is (0, 0) which makes h = 0 and k = 0.
Plugging in the values we calculated, we get:
(x-0)^2/1 - (y-0)^2/b^2 = 1
x^2 - y^2/b^2 = 1
Now we need to find the value of "b." From the asymptotes' equation y = +/-1/2x, we can find that |b/a| = 1/2. Given that a = 1, we can solve for b: |b/1| = 1/2, which gives |b| = 1/2. We can take b = 1/2.
Therefore, the final equation of the hyperbola with the given foci and asymptotes is:
x^2 - y^2/(1/2)^2 = 1
x^2 - 4y^2 = 4
The equation of the hyperbola is x^2 - 4y^2 = 4.
I use a slightly different approach to conics than the one Steve just showed you in your previous post, but the end results are the same.
I keep the values of a and b always associated with the x and y, no matter which way the hyperbola is oriented.
If it has its main axis horizontally, I use
x^2/a^2 - y^2/b^2 = 1
If it has its main axis vertically, I use
x^2/a^2 - y^2/b^2 = -1
I also sketch the box formed by the asymptotes which forms the skeleton for my graph
from y = (1/2)x I know b/a = 1/2
a = 2b
and in any hyperbola: a^2 + b^2 = c^2
(2b)^2 + b^2 = 64
5b^2 = 64
b^2 = 64/5
b = 8/√5
a = 16/√5 ----> a^2 = 256/5
equation:
x^2/(256/5) - y^2/(64/5) = -1
or
5x^2/256 - 5y^2/64 = -1