The equation `x = (2atheta)/(1+theta^2), y = (a(1-theta^2))/(1+theta^2)` where `a` is constant, is the parametric equation of the curve (A) `x^2 - y^2 = a^2` (B) `x^2 + 4y^2 = 4a^2` (C) `x^2 + y^2 = a^2` (D) `x-2y=a^2`

Find the parametric equation of the circles : x^2 + y^2 + 2x - 4y - 1 =0

The curve with parametric equations x=1 +4cos theta , y= 2 +3 sin theta . is

The curve with parametric equations x=1+4 cos theta, y=2+3 sin theta is

If x=a (theta-sin theta) and y=a(1-cos theta) , find (d^(2)y)/(dx^(2)) at theta=pi.

If x = a cos theta - b sin theta and y = a sin theta + b cos theta, prove that x^2 + y^2 = a^2 + b^2.

If x = 2sin^(2)theta and y = 2 cos^(2)theta+1 ,then find x+y

If x=theta-(1)/(theta)" and "y=theta+(1)/(theta)," then: "(d^(2)y)/(dx^(2))=

The equation of normal to the curve y=2x^(3)-x^(2)+2 at ((1)/(2),2) is a) 2x-4y+7=0 b) 2x+y-3=0 c) x+y=0 d). 2x-y+3=0

Three sides of a triangle are represented by lines whose combined equation is (2x+y-4) (xy-4x-2y+8) = 0 , then the equation of its circumcircle will be : (A) x^2 + y^2 - 2x - 4y = 0 (B) x^2 + y^2 + 2x + 4y = 0 (C) x^2 + y^2 - 2x + 4y = 0 (D) x^2 + y^2 + 2x - 4y = 0