Given $$(x ^ 2 + y ^ 2 – 3 ) ^ 3 = ( x y ) ^ 3 -x ^ 2 – y ^ 2$$
How do you find the tangent line at point $(1, 1)$ on the curve above?
I'm having trouble with this because I'm always ending up with a very long equation when I try to simplify its first derivative :c
As first derivative I have $$x(x3y^3-2) = 6x(x^2+y^2-3)^2$$
I have tried to solve this for $y$ so I can insert $x=1$ into the equation to determine the slope at $x$ which is needed for the tangent line.. buuut I haven't found any way to solve that for $y$ because the more I try to solve / simplify, the longer and more complicated the equation gets.
Maybe there is another way to do this without taking the derivative? :/
Best Answer
If You want a solution without derivative for this specific problem.
Notice that the equation is symetrical in $x$ and $y$? This means the curve intersects line $y=x$ perpendicularly i.e its tangent line at $(a,a)$ is $y+x=2a$.
For this case $a=1$ so the tangent line is $y+x=2$