How to find the global equation of a dual curve?

How to find the equation of the curve? PLEASE HELP, 10 POINTS?

• A curve has equation x^2 + 2y^2 + 5x + 6y = 10. Find the equation of the tangent to the curve at the point (2, -1). Give your answer in the form ax+by+c=0, where a, b, and c are integers. Please help me solve this problem, 10 points??

• Answer:

  x^2 + 2y^2 + 5x + 6y = 10 ----> implicit differentiation 2x + 4y * y' + 5 + 6y' = 0 4y * y' + 6y' = - 2x - 5 y' * (4y + 6) = - 2x - 5 y' = m = (-2x - 5)/(4y + 6) ---> (2 , -1) y' = m = (-2 * 2 - 5)/(4 * -1 + 6) y' = m = (-4 - 5)/(-4 + 6) y' = m = (-9/2) y = mx + b ----> m = (-9/2) & (2 , -1) -1 = (-9/2) * 2 + b -1 = -9 + b -1 + 9 = b 8 = b y = (-9/2)x + 8 <----this equation is tangent to x^2 + 2y^2 + 5x + 6y = 10 at (2, -1) http://www.wolframalpha.com/input/?i=x%5E2+%2B+2y%5E2+%2B+5x+%2B+6y+%3D+10+%26+y+%3D+%28-9%2F2%29x+%2B+8 ======= free to e-mail if have a question