The ends of minor axis are given that (-3,2) and (5,2)....

The ends of minor axis are given that (-3,2) and (5,2)....
Should we take (h , a+k) or (a + h , k) as the ends of minor axis?
Eccentricity = 3/5.what is the equation of ellipse..????

The ends of minor axis are given that (-3,2) and (5,2)
1) The eqn of the minor axis is y = 2, which shows that minor axis is parallel to the x-axis
and consequently the major axis is parallel to the y-axis.
2) The length of the minor axis = 2b = 8 => b = 4
3) e = 3/5 ; b² = a²(1 - 3²/5²) = (16/25)a² => a = (5/4)b = 5
4) The mid-point of the minor axis is {(5 - 3)/2, (2 + 2)/2} = (1, 2)
Thus the Centre of the ellipse is C (h, k) = (1, 2)
Hence the eqn. of the ellipse is {(y - 2)² / 5²} + {(x - 1)² / 4²} = 1

Is math always so tough to be hated:s

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1) In this case Minor axis of the Ellipse is parallel to x-axis, since the y-coordinates of its end points are constant. So, its Major axis is parallel to y-axis. Hence equation of the ellipse is of the form:

(x-h)²/b² + (y-k)²/a² = 1, where 2a is the measure of Major axis segment and 2b is the measure of Minor axis segment and (h,k) is the center of the ellipse.

2) ==> The end points of the minor axis are: (h+b, k) and (h-b, k)

Comparing this with the given coordinates of the end points (5,2) and (-3,2), k = 2

h+ b = 5 and h-b = -3; Solving this, h = 1 and b = 4

3) Also we have by relation, b² = a²(1 - e²), where e is the eccentricity.

==> 16 = a²(1 - 9/25); solving av = 25

4) Thus equation of the ellipse in one standard form is:

[(x-1)²/16] + [(y-2)²/25] = 1

Expanding and simplifying, 25x² + 16y² - 50x - 64y - 311 = 0