" 1."(x^(2))/(36)+(y^(2))/(16)=1

For the given ellipse ,find the equation of directrix (i) (x^(2))/(25) + (y^(2))/(9) = 1 iii (x^(2))/(36) + (y^(2))/(16) = 1 (x^(2))/(49) + (y^(2))/(36) = 1 (iv) (x^(2))/(100) + (y^(2))/(400) = 1.

The eccentricity of the ellipes (x ^(2))/(36)+ (y^(2))/(16) =1 is

The eccentricity of the ellipse (x^(2))/(36)+ (y^(2))/(16)=1 is

The eccentricity of the ellipes (x ^(2))/(36)+ (y^(2))/(16) =1 is

The eccentricity of the ellipse (x^(2))/(36 ) + (y^(2))/(16) =1 is

If P is any point on the ellipse (x^(2))/(36) + (y^(2))/(16) = 1 , and S and S' are the foci, then PS + PS' =

Find the latus rectum of ellipse (x^(2))/(36) + (y^(2))/(16) = 1 .

The eccentricity to the ellipse (x^(2))/(36)+(y^(2))/(16)=1 is

The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola (x^(2))/(36)-(y^(2))/(16)=1 is