Find the equation of an ellipse whose axes lie along the coordinates axes, which passes through the point (- 3, 1) and has eccentricity equal to .

Given that we need to find the equation of the ellipse whose eccentricity is and passes through (- 3,1).



Let us assume the equation of the ellipse is - - - - (1) (a2>b2).


We know that eccentricity of the ellipse is




5a2 - 5b2 = 2a2


5b2 = 3a2


..... - - - (2)


Substituting the point (- 3,1) in (1) we get,






5b2 = 32



From (2),




The equation of the ellipse is





3x2 + 5y2 = 32


The equation of the ellipse is 3x2 + 5y2 = 32.


15