Find the tangent and normal plane equation of the curve X ^ 2-y ^ 2 = 3 and x ^ 2 + y ^ 2-z ^ 2 = 4 at the point (- 2, - 1,1)

Find the tangent and normal plane equation of the curve X ^ 2-y ^ 2 = 3 and x ^ 2 + y ^ 2-z ^ 2 = 4 at the point (- 2, - 1,1)

X ^ 2-y ^ 2 = 3 (1) x ^ 2 + y ^ 2-z ^ 2 = 4 (2) (1) (2) derive x respectively: 2x-2ydy / DX = 02x + 2ydy / dx-2zdz / DX = 0, so: dy / DX = x / ydz / DX = 2x / Z, so dy / DX | (- 2, - 1,1) = 2dz / DX | (- 2, - 1,1) = - 4, so the tangent equation is: (x + 2) / 1 = (y + 1) / 2 = (Z-1) / - 4