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The equation of the ellipse with focus at (±5, 0) and x = 36/5 as one directrix isMarks:1
The length of the semi-latus rectum of an ellipse is one third of its major axis, its eccentricity isMarks:1
The eccentricity of the ellipse 25x 2 + 9y2 = 225 isMarks:1
The given equation is 25x2 + 9y2 = 225
x2/9 + y2/25 = 1.
Comparing with standard equation we get
a = 3, b = 5,
a = b(1 - e2)
3 = 5(1 - e2)
e2 = 1 - (9/25)
e = 4/5.
The eccentricity of the ellipse 9x2 + 5y2 – 30y = 0 isMarks:1
The given equation can be written as
9x2 + 5(y – 3)2 = 45 (x - 0)2/5 + (y - 3)2/9 = 1
Hence, a2 = 5, b2 = 9 and the eccentricity is given by
a2 = b2 (1 – e222)
5 = 9(1– e2)
Therefore, e = 2/3.
The eccentricity of an ellipse whose latus rectum is equal to distance between foci isMarks:1
The length of latus rectum of an ellipse is 2b2/a and
distance between foci is 2ae, where e is ecentricity of the ellipse.
2ae = 2b2/a (given )
e = b2/a2
Also, we know that
e2 = 1- b2/a2
e2 = 1- e
e2 + e -1 = 0