Find an equation of a the tangent line to the graph of $x^2 - xy - y^2= 1$ when $(x,y) = (2,1)$.
Here's what I have so far:
$$ x^2 - xy - y^2 = 1 $$
$$\implies x^2 - xy - y^2 - 1 = 0 $$
$$\implies y' = 2x - xy' - 2yy' $$
$$\implies -2x = y'(-x -2y) $$
By trying to simplify the equation so that $y'$ is on one side, is that the correct way to begin solving the problem in the first place? If so, is my simplification correct as shown above?
$$ x^2 - xy - y^2 = 1 $$
$$\implies x^2 - xy - y^2 - 1 = 0 $$
$$\implies y' = 2x - xy' - 2yy' $$
$$\implies -2x = y'(-x -2y) $$
Therefore, $$ y' = \frac{2x - y}{x+2y} $$
By using $x = 2$ and $y = 1$ from the point (2,1), plug the two values into $y$ to get the slope, $m$.
$$ y' = \frac{2(2) - (1)}{(2)+2(1)} $$
and $$m =\frac{3}{4}$$
By using the point slope formula, you can find the equation of the tangent line to the given graph at the point (2,1)
$$y - y_1 = m (x- x_1) $$
$$y-1 =\frac{3}{4}(x-2)$$
$$y = \frac{3}{4}x - \frac{1}{2}$$