implicit equation 2x^3-9xy+2y^3=0 The graph point (1,2). Find the equation of the tangent line to the graph of y=f(x)

Question

A function y=f(x) satisfies the implicit equation 2x^3-9xy+2y^3=0 The graph contains the point (1,2).  Find the equation of the tangent line to the graph of y=f(x)

Answer 1

y= y(x), but you do not know the function, but an implicit equation has been profided such that (x,y)= (1,2) is a solution. If you put (1,2) into the equation, you will see that indeed it is a solution. I believe the question is to find a tangent line to y(x), at the point (1,2). To do this, take the derivative of the implicit equation wrt x, remembering that you need to treat y= y(x) when taking derivatives, so that when you take the derivative of 2x^3, you will get 6x, but when you differentiate -9xy you should get -9xy' - 9y, where y' is dy/dx. Also, d/dx(2y^3) = 6y^2y'. When you do this and collect terms, you will have an equation for y'. You want to know what y' is at the point (1,2), so put in those values for x and y, and you will have the slope of y at the point (1,2), and you can now compute a line from y= mx+b at the point (1,2). I get a slope of m= 3/5 at the point (1,2). You may want to confirm. You want this slope, because that is the slope of a line that is tangent to the curve y= y(x).- Ben