Find the tangent line to the curve $x^2y – y^2 + x = 11$ at the point $(3,1)$
I tried to solve it using parametric equations
\begin{cases}
y = t \\[4px]
x = -\dfrac{1}{2t} + \dfrac{\sqrt{1+4t^3 + 44t}}{2t}
\end{cases}
and the derivative of $x(t)$, $y(t)$, $t= 1$ gives the direction vector $( 5/7 , 1 )$
so the line that passes through $(3,1)$
l: $(3,1) + s(5/7 , 1) $
Is it correct? Because using desmos the line doesn't seem to be tangent to the curve.
Best Answer
The tangent line at $(x_0,y_0)$ has equation $$ (x-x_0)\frac{\partial f}{\partial x}+ (y-y_0)\frac{\partial f}{\partial y}=0 $$ where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since \begin{align} \frac{\partial f}{\partial x}&=2xy+1 \\[6px] \frac{\partial f}{\partial y}&=x^2-2y \end{align} the tangent line has equation $$ 7(x-3)+7(y-1)=0 $$ that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields $$ x=\frac{-1\pm\sqrt{1+44t+4t^3}}{2t} $$ If $t=1$, we get $$ x=\frac{-1\pm7}{2} $$ so we need to take the branch with $+$.
The derivative of $x(t)$ is $$ x'(t)=\frac{1}{2}\cdot\frac{\dfrac{22+6t^2}{\sqrt{1+44t+4t^3}}-(\sqrt{1+44t+4t^3}-1)}{t^2} $$ For $t=1$ we get $$ x'(1)=\frac{1}{2}\frac{\dfrac{28}{7}-(7-1)}{1}=-1 $$