The curve described parametrically by x=t^2+t+1, y=t^2-t+1 represents

Step-by-step solution

We have, x=t^2+t+1 ...(i) and y=t^2-t+1 ...(ii) Now, x+y=2 (1+t^2 ) ...(iii) and x-y=2 t ...(iv) Now, from Eqs. (iii) and (iv), we get aligned & x+y=2 [1+ ( x-y 2 )^2 ] & x+y=2 [ 4+x^2+y^2-2 x y 4 ] & x^2+y^2-2 x y-2 x-2 y+4=0 ...(v) aligned On comparing with a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 We get, a=1, b=1, c=4, h=-1, g=-1 , f=-1 =a b c+2 f g h-a f^2-b g^2-c h^2 Now, =1 1 4+2(-1)(-1)(-1)-1 (-1)^2 aligned & -1 (-1)^2-4(-1)^2 & =4-2-1-1-4 & =-4 therefore, 0 aligned and a b-h^2=1 1-(1)^2=1-1=0 So, it is equation