Find the equation of the tangent line to: $x^2-xy+y-3=0$ at the point $x=-1$
I've tried to use implicit differentiation to do this, but in the answer I got, if I substitute $x=-1$ into the equation it will be undefined. I'm not sure I've done it right. Any help will be much appreciated.
On
Without resorting to implicit differentiation, $$ x^2 - xy + y - 3 = 0 $$ could be rearranged as $$ y = \frac{3-x^2}{1-x}$$ (for $x \neq 1)$ , the derivative of which equals $$ \frac{x^2 - 2x +3}{(1-x)^2} $$ The derivative evaluated at $x=-1$ equals $\frac{3}{2} $. Also, one verifies that for $x=-1$, $y=1$.
So the equation for the tangent line equals $$ y - 1 = \frac{3}{2} (x+1) $$
When $x = -1$, we have $$ (-1)^2 - (-1)y + y - 3 = 1 + y + y - 3 = -2 + 2y = 0 \implies y = 1. $$ Thus we are seeking $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $(x,y) = (-1,1)$. Via implicit differentiation, we obtain \begin{align} x^2 - xy + y - 3 = 0 &\implies 2x - \left( x\frac{\mathrm{d}y}{\mathrm{d}x} + y\right) + \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\ &\implies 2x - x\frac{\mathrm{d}x}{\mathrm{d}y} - y + \frac{\mathrm{d}x}{\mathrm{d}y} = 0 \\ &\implies (1-x)\frac{\mathrm{d}x}{\mathrm{d}y} = y-2x \\ &\implies \frac{\mathrm{d}x}{\mathrm{d}y} = \frac{y-2x}{1-x}.\tag{$\ast$} \end{align} Substituting the point $(x,y) = (-1,1)$ into ($\ast$), we conclude that $$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-2(-1)}{1-(-1)} = \frac{1+2}{1+1} = \frac{3}{2}. $$ This gives the slope of the tangent line. To obtain the equation for the tangent line, remember that a line with slope $m$ passing through a point $(x_0,y_0)$ has equation $$ y-y_0 = m(x-x_0).$$ Therefore an equation for the line tangent to the given curve through the point $(-1,1)$ is given by $$y - 1 = \frac{3}{2} (x+1). $$