The foci of hyperbola `xy = `((a^2 + b^2)/(4))` are given by (A) `x = y = +- ((a^2 + b^2)/(4a))` (B) `x = y = +- ((a^2 + b^2)/(4b))` (C) `x = y = +- ((a^2 + b^2)/(2a))` (D) `x = y = +- ((a^2 + b^2)/(2b))`

x+y=a+b , a x-b y=a^2-b^2

Solve for 'x' and 'y': (a -b) x + (a + b) y =a^2 - b^2 - 2ab and (a + b) (x + y) = a ^2+ b^2

Locus of all such points from where the tangents drawn to the ellipse x^2/a^2 + y^2/b^2 = 1 are always inclined at 45^0 is: (A) (x^2 + y^2 - a^2 - b^2)^2 = (b^2 x^2 + a^2 y^2 - 1) (B) (x^2 + y^2 - a^2 - b^2)^2 = 4(b^2 x^2 + a^2 y^2 - 1) (C) (x^2 + y^2 - a^2 - b^2)^2 = 4(a^2 x^2 + b^2 y^2 - 1) (D) none of these

{:((a^(2))/(x) - (b^(2))/(y) = 0),((a^(2)b)/(x)+(b^(2)a)/(y) = "a + b, where x, y"ne 0.):}

Solve (x)/(a) + (y)/(b) = 1 and a(x - a)- b(y + b) = 2a^(2) + b^(2) .

The locus of the midpoints of the chords of the circle x^2+y^2-a x-b y=0 which subtend a right angle at (a/2, b/2) is (a) a x+b y=0 (b) a x+b y=a^2+b^2 (c) x^2+y^2-a x-b y+(a^2+b^2)/8=0 (d) x^2+y^2-a x-b y-(a^2+b^2)/8=0

Solve the following system of equations in xa n dy a x+b y=1 b x+a y=((a+b)^2)/(a^2+b^2)-1or ,b x+a y=(2a b)/(a^2+b^2)

Simplify using following identiy: (a +- b) (a^(2) ab + b^(2)) = a^(3) +- b^(3) (2x + 3y) (4x^(2) - 6xy + 9y^(2))

Area of the quadrilateral formed with the foci of the hyperbola x^2/a^2-y^2/b^2=1 and x^2/a^2-y^2/b^2=-1 (a) 4(a^2+b^2) (b) 2(a^2+b^2) (c) (a^2+b^2) (d) 1/2(a^2+b^2)

If the foci of the ellips (x^(2))/( 25)+ ( y^(2))/( 16) =1 and the hyperbola (x^(2))/(4) - ( y^(2))/( b^(2) ) =1 coincide ,then b^(2)=