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Find the equation of latus rectum of the ellipse 9x^2 + y^2 -36 =0
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9 x^2 + y^2 = 36
9x^2/36 + y^2/36 = 1
x^2/4 + y^2/36 = 1
As we know latus rectum passes through focus and perpendicular to axis
So it's equation is y = be
As here a= 2
And b = 6
So b>a
So it becomes conjugate ellipse
So a^2 =b^2 ( 1 - e^2 ) is relation
4 = 36 ( 1 - e^2)
1/9 = 1 - e^2
e^2 = 1 - 1/9 = 8/9
e = 2√2/3
So y= be = 12√2/3 = 4√2
So required equation is
y = 4√2
✌✌✌Dhruv✌✌✌✌
9x^2/36 + y^2/36 = 1
x^2/4 + y^2/36 = 1
As we know latus rectum passes through focus and perpendicular to axis
So it's equation is y = be
As here a= 2
And b = 6
So b>a
So it becomes conjugate ellipse
So a^2 =b^2 ( 1 - e^2 ) is relation
4 = 36 ( 1 - e^2)
1/9 = 1 - e^2
e^2 = 1 - 1/9 = 8/9
e = 2√2/3
So y= be = 12√2/3 = 4√2
So required equation is
y = 4√2
✌✌✌Dhruv✌✌✌✌
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