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

Answer 1

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).

Answer 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

Answer 3

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).

Answer 4

your answer is slghtly wrong
9(x + 4)^2 + 25(y - 1)^2 = 225 not 224

Answer 5

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).