If `x^2-y^2sec^2theta=10` be a hyperbola and `x^2sec^2theta+y^2=5` be an ellipse such that the eccentricity of hyperbola=`sqrt(5)` eccentricity of ellipse then find the length of latus rectum of ellipse

Find the eccentricity and the length of latus rectum of the ellipse 4x^2+y^2=36 .

Find the eccentricity, the coordinates of the foci, and the length of the latus rectum of the ellipse 2x^(2)+3y^(2)=1 .

The eccentricity of the ellipse is (2)/(5) and the distance between the foci is 10. Find the length of the latus rectum of the ellipse.

The eccentricity of the ellipse is (2)/(5) and the distance between the foci is 10. Find the length of the latus rectum of the ellipse.

For some theta in(0,(pi)/(2)) ,if the eccentric of the hyperbola x^(2)-y^(2)sec^(2)theta=10 is sqrt(5) times the eccentricity of the ellipse, x^(2)sec^(2)theta+y^(2)=5 ,then the length of the latus rectum of the ellipse

If the eccentricity of the ellipse, x^2/(a^2+1)+y^2/(a^2+2)=1 is 1/sqrt6 then latus rectum of ellipse is

If the eccentricity of the ellipse, x^2/(a^2+1)+y^2/(a^2+2)=1 is 1/sqrt6 then latus rectum of ellipse is

Find the eccentricity, and the length of latus rectum of the hyperbola x^(2) - y^(2) = 2