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