Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the equation 9x² - 4y² - 90x - 32y = -305.Sketch the graph, and include these points and lines, along with the auxiliary rectangle.
First let's complete the square on
9x² - 4y² - 90x - 32y = -305
9(x^2 - 10x + ....) - 4(y^2 + 16t + ....) = -305
9(x^2 - 10x + 25) - 4(y^2 + 16t + 16) = -305 + 9(25) - 4(16)
9(x - 5)^2 - 4(y + 4)^2 = -144
divide by 144
(x-5)^2 / 16 - (y+4)^2 / 36 = -1
centre is (5,-4)
a = 4
b = 6
c^2 = a^2 + b^2
c = 2√13
hyperbola has vertical major axis
I assume that you can build your auxiliary rectangle around (5,-4),
drawing in your asymptotes (the diagonals), and sketching the curve.
graph
To find the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the equation 9x² - 4y² - 90x - 32y = -305, we need to rewrite the equation in the standard form for a hyperbola.
Step 1: Complete the square for both the x and y terms.
We start by moving the constant term to the right side of the equation:
9x² - 4y² - 90x - 32y + 305 = 0
Next, we group x and y terms separately and rearrange them:
(9x² - 90x) - (4y² + 32y) = -305
Now, let's complete the square for the x and y terms:
For the x terms:
9(x² - 10x) + 4y² + 32y = -305
To complete the square for x, we need to add and subtract half of the coefficient of x, squared. In this case, half of the coefficient of x is -10/2 = -5. Squaring this gives 25. So, we add 25 inside the parentheses and also subtract it outside the parentheses:
9(x² - 10x + 25) - 25 + 4y² + 32y = -305
Simplifying further:
9(x - 5)² + 4y² + 32y - 25 - 305 = 0
9(x - 5)² + 4y² + 32y - 330 = 0
Similarly, for the y terms:
9(x - 5)² + 4(y² + 8y) - 330 = 0
9(x - 5)² + 4(y² + 8y + 16) - 64 - 330 = 0
9(x - 5)² + 4(y + 4)² - 394 = 0
Step 2: Divide both sides by -394 to get the equation in standard form with a positive constant on the right side.
(x - 5)²/(-394/9) + (y + 4)²/(-394/4) = 1
Simplifying further:
(x - 5)²/(-394/9) - (y + 4)²/(394/4) = 1
Step 3: Compare the standard form equation to the general form of a hyperbola.
The standard form of a hyperbola with center (h, k) is:
(x - h)²/a² - (y - k)²/b² = 1
From the standard form equation, we can determine the values of a² and b²:
a² = -394/9
b² = 394/4
Step 4: Find the values of a, b, center, foci, vertices, and asymptotes.
From the values of a² and b², we can calculate a and b:
a = √(-394/9)
b = √(394/4)
The center of the hyperbola is the point (h, k), which in this case is (5, -4).
The formula for finding the foci is given by c = √(a² + b²), where c represents the distance from the center to the foci.
So, c = √((-394/9) + (394/4))
The vertices can be found by adding and subtracting a from the center (h, k). In this case, (5, -4) + a and (5, -4) - a.
The asymptotes have equations y = ±(b/a)(x - h) + k. Substituting the values of a, b, h, and k into these equations will give the equations of the asymptotes.
Step 5: Sketch the graph and include the points, lines, and auxiliary rectangle.
Unfortunately, I cannot sketch the graph for you as I am a text-based AI. However, using the information provided above, you can plot the points (center, foci, vertices) on a graph, draw the asymptotes, and sketch the auxiliary rectangle.
I hope this helps you understand how to find the coordinates of the center, foci, vertices, and asymptotes of a hyperbola.
To find the coordinates of the center, foci, vertices, and asymptotes of a hyperbola, we need to rewrite the given equation in the standard form:
1. Rearrange the terms to isolate the squared terms:
9x² - 4y² - 90x - 32y = -305
9x² - 90x - 4y² - 32y = -305
2. Complete the square for both the x and y terms:
9(x² - 10x) - 4(y² + 8y) = -305
9(x² - 10x + 25) - 4(y² + 8y + 16) = -305 + 9 * 25 - 4 * 16
9(x - 5)² - 4(y + 4)² = -305 + 225 - 64
9(x - 5)² - 4(y + 4)² = -144
Now that we have the equation in standard form, we can determine the properties of the hyperbola:
1. Center:
The center of the hyperbola is represented by the values (h, k) within the equation. In our case, (h, k) = (5, -4).
2. Foci:
To find the foci, we need to use the equation c² = a² + b², where a is the distance between the center and vertices, b is the distance between the center and co-vertices, and c is the distance between the center and foci.
The value of a can be determined by taking the square root of the coefficient in front of the (x - h)² term, which in our case is 3. So a = √3.
The value of b can be determined by taking the square root of the coefficient in front of the (y - k)² term, which in our case is 2. So b = √2.
Now, we can find c using the equation c² = a² + b²:
c² = 3² + 2²
c² = 9 + 4
c² = 13
c = √13
The foci are located at a distance of c units from the center in the x-direction. So the coordinates of the foci are (5 ± √13, -4).
3. Vertices:
The vertices lie on the transverse axis and are located at a distance of a units from the center in the x-direction. So the coordinates of the vertices are (5 ± √3, -4).
4. Asymptotes:
The equations of the asymptotes in the standard form are y = ±(b/a)(x - h) + k. Substituting the values, we get the equations:
y = ±(√2/√3)(x - 5) - 4
Now, let's sketch the graph of the hyperbola:
1. Draw the coordinate axes.
2. Draw the asymptotes using the equations we found: y = (√2/√3)(x - 5) - 4 and y = - (√2/√3)(x - 5) - 4.
3. Plot the center at (5, -4).
4. Plot the foci at (5 + √13, -4) and (5 - √13, -4).
5. Plot the vertices at (5 + √3, -4) and (5 - √3, -4).
6. Sketch the hyperbola by drawing a curve that approaches but does not touch the asymptotes, passing through the vertices and center.
7. Draw an auxiliary rectangle around the hyperbola, aligning its sides to the major and minor axes.
Note: It's important to remember that accurate graphing may require using a scale appropriate for the values involved.