Find the foci of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth. 9x^2 + 25y^2 - 54x + 150y + 81 =0
To find the foci of the ellipse given by the equation 9x^2 + 25y^2 - 54x + 150y + 81 = 0, we first need to put it into standard form by completing the square:
9(x^2 - 6x + 9) + 25(y^2 + 6y + 15) - 81 = 0
Dividing everything by 225, we get:
(x^2 - 6x + 9)/25 + (y^2 + 6y + 15)/9 = 1
Now we can see that this is an ellipse centered at (3, -2) with a^2 = 25 and b^2 = 9, where a is the distance from the center to the horizontal vertices and b is the distance from the center to the vertical vertices.
The distance from the center to the foci is given by c = sqrt(a^2 - b^2). Plugging in our values, we get:
c = sqrt(25 - 9) = sqrt(16) = 4
Therefore, the foci of the ellipse are located 4 units to either side of the center, or at (3 + 4, -2) = (7, -2) and (3 - 4, -2) = (-1, -2).
So the foci of the ellipse are (7, -2) and (-1, -2).
Find the vertices of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth. 9x^2 + 25y^2 + 72x - 50y - 56 = 0
To find the vertices of the ellipse given by the equation 9x^2 + 25y^2 + 72x - 50y - 56 = 0, we need to put it into standard form by completing the square:
9(x^2 + 8x + 16) + 25(y^2 - 2y + 1) - 56 - 144 - 25 = 0
Simplifying, we get:
9(x + 4)^2 + 25(y - 1)^2 = 224
Dividing everything by 224, we get:
(x + 4)^2/24.89 + (y - 1)^2/8.96 = 1
Now we can see that this is an ellipse centered at (-4, 1) with a^2 = 24.89 and b^2 = 8.96, where a is the distance from the center to the horizontal vertices and b is the distance from the center to the vertical vertices.
The vertices are located a distance of a units to either side of the center, and b units above and below the center. Plugging in our values, we get:
Vertex on the left: (-4 - sqrt(24.89), 1) ≈ (-8.4, 1)
Vertex on the right: (-4 + sqrt(24.89), 1) ≈ (0.4, 1)
Vertex on top: (-4, 1 + sqrt(8.96)) ≈ (-4, 4)
Vertex on bottom: (-4, 1 - sqrt(8.96)) ≈ (-4, -2)
So the vertices of the ellipse are approximately (-8.4, 1), (0.4, 1), (-4, 4), and (-4, -2).
your answer is slghtly wrong
9(x + 4)^2 + 25(y - 1)^2 = 225 not 224
You are correct! Thank you for pointing out the mistake.
Using the correct value, we get:
(x + 4)^2/25 + (y - 1)^2/9 = 1
Now we can see that this is an ellipse centered at (-4, 1) with a^2 = 25 and b^2 = 9.
The vertices are located a distance of a units to either side of the center, and b units above and below the center. Plugging in our values, we get:
Vertex on the left: (-4 - 5, 1) = (-9, 1)
Vertex on the right: (-4 + 5, 1) = (1, 1)
Vertex on top: (-4, 1 + 3) = (-4, 4)
Vertex on bottom: (-4, 1 - 3) = (-4, -2)
So the vertices of the ellipse are (-9, 1), (1, 1), (-4, 4), and (-4, -2).
To find the foci of an ellipse, we need to put the given equation in standard form. The standard form of an ellipse is given by:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
Where (h, k) is the center of the ellipse, "a" represents the semi-major axis, and "b" represents the semi-minor axis.
Let's start by rearranging the given equation into standard form:
9x^2 - 54x + 25y^2 + 150y + 81 = 0
Group the x and y terms separately:
9x^2 - 54x + 25y^2 + 150y = -81
Now, complete the square for both x and y terms. Let's start with the x terms:
9(x^2 - 6x) + 25y^2 + 150y = -81
To complete the square for the x terms, we need to take half of the coefficient of x (-6 in this case) and square it, then add it to both sides of the equation:
9(x^2 - 6x + 9) + 25y^2 + 150y = -81 + 9 * 9
9(x - 3)^2 + 25y^2 + 150y = -81 + 81
9(x - 3)^2 + 25y^2 + 150y = 0
Similarly, complete the square for the y terms:
9(x - 3)^2 + 25(y^2 + 6y) = 0
9(x - 3)^2 + 25(y^2 + 6y + 9) = 0 + 25 * 9
9(x - 3)^2 + 25(y + 3)^2 = 225
Now, divide through by 225 to get the equation in standard form:
((x - 3)^2) / (25/9) + ((y + 3)^2) / (9/25) = 1
Comparing this with the standard form equation, we can identify the center:
Center: (h, k) = (3, -3)
Next, we can find the values of a and b by comparing the denominators:
a^2 = 25/9
a = √(25/9) = 5/3
b^2 = 9/25
b = √(9/25) = 3/5
The distance between the center and each focus is determined by the value of c, where c^2 = a^2 - b^2:
c^2 = (5/3)^2 - (3/5)^2
c^2 = 25/9 - 9/25
c^2 = (625 - 81) / (9 * 25)
c^2 = 544 / 225
c = √(544/225)
To find the foci, we need to add or subtract c from the x-coordinate of the center:
Foci: (3 ± (c), -3)
Calculating the value of c, rounded to the nearest tenth:
c ≈ √(544/225) ≈ 1.19
Therefore, the foci of the ellipse are approximately:
Foci: (3 ± 1.19, -3) ≈ (4.19, -3) and (1.81, -3)
To find the foci of an ellipse, we need to express the equation of the ellipse in standard form:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Where (h, k) are the coordinates of the center, and 'a' and 'b' represent the semi-major and semi-minor axes of the ellipse.
Let's rearrange the given equation to match the standard form:
9x^2 - 54x + 25y^2 + 150y = -81
Start by completing the square for the x-terms:
9(x^2 - 6x) + 25y^2 + 150y = -81
To complete the square for x, take half of the coefficient of 'x', square it, and add it within the parentheses:
9(x^2 - 6x + 9) + 25y^2 + 150y = -81 + 9 * 9
Simplify and rearrange the equation:
9(x - 3)^2 + 25y^2 + 150y = 0
Next, complete the square for the y-terms. Take half of the coefficient of 'y', square it, and add it to both sides of the equation:
9(x - 3)^2 + 25(y^2 + 6y) = 0 - 25 * 9
9(x - 3)^2 + 25(y^2 + 6y + 9) = - 225
Simplify and rearrange the equation:
9(x - 3)^2 + 25(y + 3)^2 = -225 + 25 * 9
9(x - 3)^2 + 25(y + 3)^2 = -225 + 225
9(x - 3)^2 + 25(y + 3)^2 = 0
Now we can write the equation in standard form:
(x - 3)^2/25 + (y + 3)^2/9 = 1
Comparing this to the standard form equation, we can see that the center of the ellipse is located at the point (3, -3), and the semi-major axis 'a' is √25 = 5, while the semi-minor axis 'b' is √9 = 3.
To find the foci of the ellipse, we use the equation:
c^2 = a^2 - b^2
Let's calculate c:
c^2 = 5^2 - 3^2
c^2 = 25 - 9
c^2 = 16
c = √16
c = 4
The foci of the ellipse are located at (h ± c, k). Plugging in the values:
Foci = (3 ± 4, -3)
Foci = (7, -3) and (-1, -3)
Therefore, the foci of the ellipse defined by the equation 9x^2 + 25y^2 - 54x + 150y + 81 = 0 are at (7, -3) and (-1, -3).