SOLUTION: Find the slope of the curve x^2+2y^2-3x-4y+2=0 at (1,2)

First, you should check that the given point (1,2) lies on the curve,
i.e. satisfies the equation.


It does satisfy, so this part is Ok.


Next, differentiate implicitly

    2x*dx + 4y*dy -3*dx - 4*dy = 0.    (1)


Here dx and dy are linear elements; x and y are parameters (coordinate values of points on the curve).


In equation (1), substitute the values  x= 1, y= 2 for the given point. You will get then

    2*1*dx + 4*2*dy - 3*dx - 4*dy = 0,  or


   2*dx    + 8*dy   - 3*dx - 4*dy,

    -dx    + 4*dy = 0

    4*dy          = dx

       = 


ANSWER.  The slope of the tangent to the curve  x^2+2y^2-3x-4y+2=0  at  (1,2)  is  .