Let x^2/a^2+y^2/b^2=1(agtb) be a given ...

Let `x^2/a^2+y^2/b^2=1(agtb)` be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, `pphi(t)=5/2+t-t^2`, then `a^2+b^2` is equal to :

Similar Questions

Explore conceptually related problems

Find the equation of an ellipse whose latus rectum is 10 and eccentricity is (1)/(2) .

If the length of latus rectum is 5/2 and eccentricity is 1/2 , then the equation of the ellipse is

Find equation of ellipse whose l(latus-rectum) = (5)/(2) and eccentricity y = (1)/(2)

For an ellipse (x^2)/(a^2) +(y^2)/(b^2) =1, a gt b , the area enclosed by two latus rectum is ……. ( where is the eccentricity of an ellipse )

Find the latus rectum , eccentricity and the coordinates of the foci of the ellipse 9x^(2) + 5y^(2) + 30 y = 0

An ellipse has directrix x+y=-2 focus at (3,4) eccentricity =1/2, then length of latus rectum is

An ellipse has directrix x+y=-2 focus at (3,4) eccentricity =1//2, then length of latus rectum is

Find the coordinates of the vertices, length of the latus rectum and eccentricity of the ellipse x^(2)/49 + y^(2)/36 = 1

(2,4 ) and (10 , 10 ) are the ends of the latus rectum of an ellipse with eccentricity 1/2 . The length of major axis is

Find the latus rectum, the eccentricity and coordinates of the foci of the ellipse 9x^2 + 5y^2 + 30y=0 .