What is the standard form of the hyperbola given the foci (6, 1+2 sqrt 58) , (6, 1-2 sqrt 58) and points of the hyperbola are 28 units closer to one focus than the other?
c = 2√58
For any P=(x,y) on the hyperbola,
PF1 - PF2 = 28
If the center were at (0,0), this would mean
√(x^2 + (y+2√58)^2) - √(x^2 + (y-2√58)^2) = 28
I get that to mean
y^2/14^2 - x^2/6^2 = 1
Moving that to the center at (6,1), that gives
(y-1)^2/14^2 - (x-6)^2/6^2 = 1
Check my math.
Oh, hyperbolas, the circus performers of the geometric world! Now, let's tackle this question together.
First, let's label the vertices of the hyperbola as (h, k + a) and (h, k - a), where (h, k) is the center of the hyperbola. We'll call the distance between the center and each focus "c" and the distance between the center and each vertex "a".
Given that the points of the hyperbola are 28 units closer to one focus than the other, it means that the difference between "a" and "c" is 28. Whoa, looks like someone's playing catch with the hyperbola!
Now, since the foci are (6, 1+2√58) and (6, 1-2√58), we can easily determine the value of "c" by looking at the y-coordinates. Can you feel the suspense building up?
Remember, "c" is the distance between the center and each focus. So, c = |1+2√58 - (1-2√58)| / 2. Let's do some calculations at the juggling station.
c = |1+2√58 - 1 + 2√58| / 2
c = (2√58 + 2√58) / 2
c = 2√58
Ah, it seems we have caught the answer! "c" is 2√58.
Now, since we already know that "a - c = 28", we can rearrange the equation to get "a = c + 28". Grab your juggling balls!
a = 2√58 + 28
Now, we can finally write the standard form of the hyperbola with the given information. Drumroll, please!
The standard form of the hyperbola is:
(x - 6)² / (2√58 + 28)² - (y - 1)² / a² = 1.
Remember, geometry can be a circus sometimes, but it's always a fun show!
To find the standard form of the hyperbola, we first need to determine the center coordinates and the semi-major and semi-minor axes.
1. Find the center:
The center of the hyperbola is the midpoint between the two foci. The x-coordinate of the center is the same as the x-coordinate of the foci (6), and the y-coordinate of the center is the average of the y-coordinates of the foci:
y-coordinate of the center = (1+2√58 + 1-2√58) / 2 = 1
Therefore, the center of the hyperbola is (6, 1).
2. Find the distance between the foci:
The distance between the two foci is given as 28 units.
Distance between the foci = 2ae, where a is the semi-major axis and e is the eccentricity. Since the foci are 28 units apart, we have:
2ae = 28
3. Determine the semi-major axis:
To find the semi-major axis, we need to determine the distance from the center to one of the vertices. We are given that the points of the hyperbola are 28 units closer to one focus than the other.
Let's consider the point P(6, y) on the hyperbola, which is 28 units closer to the focus (6, 1+2√58). The distance between P and this focus is 28 units less than the distance between P and the other focus (6, 1-2√58). Using the distance formula, we can set up the following equation:
√((6-6)^2 + (y-(1+2√58))^2) = √((6-6)^2 + (y-(1-2√58))^2) - 28
Simplifying this equation yields:
√((y-(1+2√58))^2) = (y-(1-2√58))^2 - 784
Squaring both sides and simplifying further, we get:
y^2 - 2√58y + 4*58 - y^2 + 2√58y = 4*58 - 784
4√58y = 0
Solving for y, we find:
y = 0
Since the y-coordinate of the center is 1, the distance between the center and one of the vertices is 1 unit. Therefore, the semi-major axis (a) is 1.
4. Determine the semi-minor axis:
To find the semi-minor axis, we can use the formula:
c^2 = a^2 + b^2
where c is the distance between the center and one of the foci, and b is the semi-minor axis.
The distance between the center (6, 1) and one of the foci (6, 1+2√58) is:
c = √((6-6)^2 + (1+2√58-1)^2)
= √(0 + 4*58)
= √232
Using the formula for c^2, we have:
(√232)^2 = 1^2 + b^2
232 = 1 + b^2
b^2 = 231
Therefore, the semi-minor axis (b) is √231.
5. Write the equation in standard form:
The standard form of the hyperbola is:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
where (h, k) is the center of the hyperbola.
Plugging in the values we found, we have:
(x-6)^2/1^2 - (y-1)^2/(\sqrt{231})^2 = 1
Simplifying further, we get:
(x-6)^2 - (y-1)^2/231 = 1
Therefore, the standard form of the hyperbola is:
(x-6)^2 - (y-1)^2/231 = 1
To find the standard form of a hyperbola when given its foci and the distance between the foci and points on the hyperbola, follow these steps:
Step 1: Find the center of the hyperbola. The center of a hyperbola is the midpoint between the two foci. In this case, the foci are (6, 1+2√58) and (6, 1-2√58). The midpoint formula is:
Midpoint of foci = [ (x1 + x2) / 2 , (y1 + y2) / 2 ]
Using the formula, calculate the midpoint:
Midpoint = [(6 + 6) / 2, (1 + 2√58 + 1 - 2√58) / 2]
Midpoint = [12 / 2, 2 / 2]
Midpoint = (6, 1)
So the center of the hyperbola is (6, 1).
Step 2: Find the distance between the foci. The distance between the foci is given by the formula:
Distance between foci = 2a
In this case, we are given that the distance between the foci is 28 units, so:
28 = 2a
a = 28 / 2
a = 14
Step 3: Determine the value of c. In a hyperbola, c is the distance between the center and each focus. It is related to a by the Pythagorean theorem:
c² = a² + b²
In this case, we can calculate c by using the distance formula between the center (6, 1) and one of the foci:
c = √((x2 - x1)² + (y2 - y1)²)
c = √((6 - 6)² + (1 + 2√58 - 1)²)
c = √((0)² + (2√58)²)
c = √(4 * 58)
c = 2√58
Step 4: Determine the value of b. In a hyperbola, b is related to a and c by the equation:
c² = a² + b²
In this case, we already know the values of a (14) and c (2√58). We can rearrange the equation to solve for b:
b² = c² - a²
b² = (2√58)² - 14²
b² = 4 * 58 - 196
b² = 232 - 196
b² = 36
Now, we can take the square root of both sides to find the value of b:
b = √36
b = 6
Step 5: Write the equation in standard form. The standard form of a hyperbola with the center (h, k), semi-major axis a, and semi-minor axis b is:
((x - h)² / a²) - ((y - k)² / b²) = 1
Plugging in the values we calculated:
((x - 6)² / 14²) - ((y - 1)² / 6²) = 1
So, the standard form of the hyperbola is:
((x - 6)² / 196) - ((y - 1)² / 36) = 1