Find the endpoints of the minor axis:?

Question

9x^2 + 16y^2 - 54x - 64y + 1 = 0

Answer 1

You need to put this equation in standard form.

9x² + 16y² - 54x - 64y + 1 = 0
9(x² - 6x) + 16(y² - 4y) = -1

Complete the squares on the left side:

9(x² - 6x + 9) + 16(y² - 4y + 4) = -1 + 81 + 64
9(x - 3)² + 16(y - 2)² = 144

The standard form has 1 on the right side, so divide both sides by 144:

[(x - 3)²]/16 + [(y - 2)²]/9 = 1

This is in standard form. We read off that the center is at (3,2), the semi-major axis (which is parallel to the x-axis) is 4, and the semi-minor axis (which is parallel to the y-axis) is 3. So the endpoints of the minor axis are 3 units above and below the center.

Answer 2

You've got to complete squares:
9(x^2 -6x +9) + 16(y^2 -4y +4) = -1 +81 + 64
9(x-3)^2 +16(y - 2)^2 = 12^2
You need a 1 on the right side and a 1 in front of both (.
(x -3)^2/4^2 +(y - 2)^2/3^2 =1
The center of the ellipse is (3, 2). The length of the minor axis is 2*3 (the sqrt of the smaller of the denoms. *2)
So, (3, 2 +3) and (3, 2 - 3) are the culprits.
Check all my arithmetic!

Answer 3

that could help
http://www.mathwords.com/h/hyperbola.htm