Thread: This Alg 2 problem is hard

Find the equation of the ellipse specified:

(a) foci (-2,1) (4,1) and minor axis 8

(b) vertices (+-4,0) (0,+-5)

can you explain how to do it?

Also this extra credit problem is getting me frustrated.

sketch the graph of the equation and indicate the center, the foci, and the length of axes.

2x squared + 3y squared + 12x -24y + 60= 0

do i need to group it and then factor? what would my next steps be?? thanks

Originally Posted by mattballer082

Find the equation of the ellipse specified:

(a) foci (-2,1) (4,1) and minor axis 8

Hello,

the center of the ellipse is the midpoint between the foci: C(1, 1)
I assume that the axes of the ellipse are parallel to the coordinate axes, otherwise your problem can't be done:
Let a be the major axis and let be b the minor axis.
You know that the distance between center and the focus on the major axis is called excentricity e and that it is calculated by:
e² = a² - b² . Thus e = 4 - 1 = 3

e² + b² = a². With your problem:
9 + 64 = 73 = a². a = √(73)

Therefore the equation of your ellipse becomes:
x²/73 + y²/64 = 1

Originally Posted by mattballer082

(b) vertices (+-4,0) (0,+-5)

can you explain how to do it?

The midpoint of all vertices is the center of the ellips: C(0, 0)
The length of the axes is the distance between center and vertices. Therefore
a = 4, b = 5

The equation of the ellipse becomes:

x²/16 + y²/25 = 1

Originally Posted by mattballer082

Also this extra credit problem is getting me frustrated.

sketch the graph of the equation and indicate the center, the foci, and the length of axes.

2x squared + 3y squared + 12x -24y + 60= 0
...

Firts: Complete the squares:

2x² + 3y² + 12x - 24y + 60 = 0
2(x² + 6x + 9 - 9) +3(y² - 8y + 16 - 16) = -60
2(x+3)² - 18 + 3(y-4)² - 48 = -60
2(x+3)² + 3(y-4)² = 6
(x+3)²/3 + (y-4)²/2 = 1

The center of the ellipse is C(-3, 4)

The major axis is a = √3; the minor axis is b = √2

e² = a² - b² ==> e² = 3 - 2 = 1

Therefore F1(3+1, 4) and F2(3-1, 4)