The equation of the ellipse in the form of `x^2/a^2+y^2/b^2=1`, given the eccentricity to be `2/3` and latus recturn `2/3`, is

Find the equation of the ellipse in the form x^2/a^2+y^2/b^2=1(a>b) , given : length of the LR is 5/2 and eccentricity is sqrt3/2

Find the equation of the ellipse in the form x^2/a^2+y^2/b^2=1(a>b) , given : major axis is 2sqrt5 , eccentricity is 1/sqrt2

Find the equation of the ellipse in the form x^2/a^2+y^2/b^2=1(a>b) , given : one of the foci is (3,0), eccentricity is 3/5 .

Find the equation of the ellipse in the form ((x-h)^2)/a^2+((y-k)^2)/b^2=1(a>b) , given : vertices (2,-2) and (2,-4) and eccentricity is 1/3 and distance between the foci is 4.

An equation of the normal to the ellipse x^2//a^2+y^2//b^2=1 with eccentricity e at the positive end of the latus rectum is

Find the equation of the ellipse in the form ((x-h)^2)/a^2+((y-k)^2)/b^2=1 . Given the following data. Centre(2,-1) , e= 1/2 , length of latus rectum 4.

Find the equation of the ellipse in the form x^2/a^2+y^2/b^2=1(a>b) , given : ellipse passes through the points (2,3) and (-4,1)

Find the equation of the normal to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the positive end of the latus rectum.

Find the equation of the normal to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the positive end of the latus rectum.

Find the equation of the normal to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the positive end of the latus rectum.