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Grade: 12th pass

                        

Find the equation to the ellipse whose one vertex is (3,1),the nearer focus is (1,1) and the eccentricity is 2/3.

2 years ago

Answers : (1)

Arun
24742 Points
							
Dear Rahul
 
Consider x^2/a^2+y^2/b^2=1 
the distance from the vertex to the focus is 2 
therefore, a-ae=2 
but e=2/3 and thus a=6 

We also know 
b^2=a^2(1-e^2) 
b^2=6^2(1-2^3/3^2) 
b^2=20 
Thus if the centre is at the origin, the ellipse would be x^2/36+y^2/20=1 
But your ellipse is shifted to the left by 3 and up by 1 therefore, your ellipse's equation is 
(x+3)^2/36+(y-1)^2/20=1
2 years ago
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