"x^2-xy+y^2=3" is the equation of an ellipse. By implicit differentiation determine the equation of the normal of the given equation at (-1,1)
Let "F(x,y)=x^2-xy+y^2-3." Then "F_x=2x-y" and "F_y=-x+2y," and hence "(F_x(-1,1),F_y(-1,1))=(-3,3)" is a normal vector to the curve at the point "(-1,1)." It follows that the equation of the normal at "(-1,1)" is "\\frac{x+1}{-3}=\\frac{y-1}{3}," which is equivalent to "y=-x."
Comments
Leave a comment