Curve D has equation 3x^2+2xy-2y^2+4=0 Find the equation of the tangent at point (2,4) and give your answer in the form ax+by+c=0, were a,b and c are integers.

Differentiate the equationdy/dx=6x+2x(dy/dx)+2y-4y(dy/dx)Set this equal to zero and solve for dy/dx which gives:dy/dx=(2y+6x)/(4y-2x)For x=2 and y=4 dy/dx=5/3(y-y_o)=m(x-x_o)y-4=(5/3)(x-2)Thus, the equation of the tangent at (2,4) is 5x-3y+2=0