Find the equation of the ellipse whose foci are (-2,6) and (-2,-2) and a minor axis endpoint is (0,2)?

Question

thank you

Answer 1

The center of the ellipse is (h,k) = (-2,2), determined by taking the midpoint of the two foci. The focal length, c, is half the distance between the two foci; the distance between the foci is 8, so c = 4. The point (0,2) is 2 to the right of the foci, so the semi-minor axis has length b = 2 and is horizontal. The semi-major axis length, a, is equal to the length of the hypotenuse of a right triangle formed by b and c, so a = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5), and the semi-major axis is vertical.

The equation of an ellipse is ((x - h)^2)/b^2 + ((y - k)^2)/a^2 = 1. h, k, a, and b are all calculated above. It should be noted that b and a may be interchangeable. a is usually the semi-major axis and b is the semi-minor axis. If the semi-major axis had been horizontal, a would go in the denominator of the fraction using x. Because the semi-major axis is vertical in this figure, a goes with y.

Answer 2

The equation of the unique ellipse could be re-written interior the common style x²/a² + y²/b² = a million 4x² + 9y² = 36 (4/36)x² + (9/36)y² = a million x²/3² + y²/2² = a million.....the main axis of this ellipse for that reason coincides with the x - axis. The foci are at (-c,0) and (c,0) the place c = ?(a² - b²) = ?(9 - 4) = ?5 the recent ellipse has its substantial axis coinciding with the y-axis with its center on the beginning place. via fact the size of the main axis is 10, then the vertices are at (0.5) and (0,-5) The vertices for the minor axis are time-commemorated as they are the foci of the unique ellipse. (-?5,0) and (?5,0) The equation for the recent ellipse is for that reason, y²/5² + x²/(?5)² = a million or y²/5² + x²/5 = a million. to make certain the foci we could resolve, c = ?(a² - b²) = ?(25 - 5) = ?20 = 2?5. Then the foci are at (0,2?5) and (0,-2?5)

Answer 3

Center: ((-2+-2)/2, (6+-2)/2) = (-2,2)
c = (6--2)/2 = 4
b = 0--2 = 2
a^2 = c^2 - b^2 = 16 - 4 = 12

Therefore, the equation is,
(x+2)^2/12 + (y-2)^2/4 = 1