Find the tangent and normal plane equation of curve y ^ 2 = 4x, z = 2x ^ 2 at point x = 1 I don't know whether to solve the plane equations of the two equations separately or whether the two equations are the same equation?

Find the tangent and normal plane equation of curve y ^ 2 = 4x, z = 2x ^ 2 at point x = 1 I don't know whether to solve the plane equations of the two equations separately or whether the two equations are the same equation?

Find the tangent and normal plane equation of the curve Y & # 178; = 4x, z = 2x & # 178; at the point XO = 1
Let x = t, Y & # 178; = 4T, z = 2T & # 178;; to = 1, XO = 1, yo = ± 2, Zo = 2;
DX / dt = x '= 1; 2Y (dy / DT) = 2yy' = 4, so y '= 4 / 2Y; DZ / dt = Z' = 4T;
So when to = 1, x'o = 1, y'o = ± 1, z'o = 4;
So the tangent equation of the curve at XO = 1 is: X-1 = (Y-2) / 4 = (Z-2) / 4; or X-1 = - (y + 2) = (Z-2) / 4;
The normal plane equation of the curve at XO = 1 is: (x-1) + (Y-2) + 4 (Z-2) = x + y + 4z-5 = 0, or (x-1) - (y + 2) + 4 (Z-2) = X-Y + 4z-5 = 0