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find equation of the ellipse whose one vertex is 3,1 and nearer focus is 1,1 and eccentricity 2/3. |
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Answer» Consider x^2/a^2+y^2/b^2=1the distance from the vertex to the focus is 2therefore, a-ae=2but e=2/3 and thus a=6 We also knowb^2=a^2(1-e^2)b^2=6^2(1-2^3/3^2)b^2=20Thus if the centre is at the origin, the ellipse would be x^2/36+y^2/20=1But 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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