(v)(x^(2))/(a^(2))+(y^(2))/(b^(2))=1" and "x^(2)+y^(2)=ab

The angle of intersection of the curves (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and x^(2)+y^(2)=ab, is

An angle of intersection of the curves , (x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1 and x^(2) + y^(2) = ab, a gt b , is :

ab(x^(2)+y^(2))-xy(a^(2)+b^(2))

If a tangent of slope 2 of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is the normal to the circle x^(2)+y^(2)+4x+1=0 then the maximum value of a.b is equal to

Solve the system of equations for x and y:(b^(2)x)/(a)-(a^(2)y)/(b)=ab(a+b) and b^(2)x-a^(2)y=2a^(2)b

If (x+1)/(x-1)=(a)/(b) and (1-y)/(1+y)=(b)/(a), then the value of (x-y)/(1+xy) is (2ab)/(a^(2)-b^(2)) (b) (a^(2)-b^(2))/(2ab) (c) (a^(2)+b^(2))/(2ab) (d) (a^(2)-b^(2)backslash)/(ab)

(a)/(x)-(b)/(y)=0,(ab^(2))/(x)+(a^(2)b)/(y)=a^(2)+b^(2)

a(y+z)=x,b(z+x)=y,c(x+y)=z prove that (x^(2))/(a(1-bc))=(y^(2))/(b(1-ca))=(z^(2))/(c(1-ab))