x^(2)+y^(2)=1

Find the ratio of the eccentricities of the hyperbolas 2x^(2) - 3y^(2) = 1 and 3x^(2) - 2y^(2) = 1

2x+3y=2,x-(y)/(2)=(1)/(2) .

A circle of constant radius a passes through the origin O and cuts the axes of coordinates at points P and Q . Then the equation of the locus of the foot of perpendicular from O to P Q is (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)(1/(x^2)+1/(y^2))=a^2

(x^(2^(n-1))+y^(2^(n-1)))(x^(2^(n-1))-y^(2^(n-1)))=

If X+Y=[[2,-1],[1,3],[-3,2]] and X-2Y=[[-2,-1],[-3,-1],[4,-2]] find X and Y.