364()/((c)2y^(2)=x^(3)" और "y^(2)=32x)quad " XII,Jणित भाग-१ "

Prove that the curves 2y^(2)=x^(3) and y^(2)=32x cut each other orthogonally at the origin

Find the angle of intersection of the curves 2y^(2) = x^(3) and y^(2) =32x .

Divide 14x^(3)y^(2)+8x^(2)y^(3)-32x^(2)y^(5) by -2xy^(2)

If (1 - y) (1 + 2x + 4x^(2) + 8x^(3) + 16x^(4) + 32x^(5) ) = 1 -y^(6), (y ne 1) , then a value of y/x is

Find the angle between the curves 2y^(2)=x^(3) and y^(2)=32x

Then equation of auxiliary circle of the ellipse 16x^(2)+25y^(2)+32x-100y=284 is (A)x^(2)+y^(2)+2x-4y-20=0 (B) x^(2)+y^(2)+2x-4y=0 (C) (x+1)^(2)+(y-2)^(2)=400(D)(x+1)^(2)+(y-2)^(2)=225

The equation of the image of the circle x^(2)+y^(2)+16x-24y+183=0 in the mirror 4x+7y+13=0 is x^(2)+y^(2)+32x-4y+235=0x^(2)+y^(2)+32x+4y-235=0x^(2)+y^(2)+32x-4y-235=0x^(2)+y^(2)+32x+4y-235=0x^(2)+y^(2)+32x+4y+235=0

The equatioin of the image of the circle x^(2)+y^(2)+016x-24y+183=0 in the line mirror 4x+7y+13=0 is: a.x^(2)+y^(2)+32x-4y+235=0 b.x^(2)+y^(2)+32x-4y-235=0cx^(2)+y^(2)+32x-4y-235=0dx^(2)+y^(2)+32x-4y-235=0dx^(2)+y^(2)+32x+4y+235=0