How find tangent line of the given curve at this point?

Question

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? :/

Answer 1

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 $y+x=2$

Answer 2

Your derivative is not computed properly. You need to differentiate both sides implicitly with respect to $x$ and use chain and product rules.

Here, I'm representing the first derivative (general form) as $y'$:

$3(x^2 + y^2 - 3)^2(2x + 2y\cdot y') = 3(xy)^2(y + x\cdot y') - 2x - 2y\cdot y'$

Now put in $x=y=1$ to find $y'(1)$, it's a simple linear equation.

That gives you the slope of the tangent line. The line passes through $(1,1)$. Hence find the equation. You should be able to handle this easily.

Answer 3

Let $y'(1,1)=m$.

Thus, after taking derivative of the both sides we obtain: $$3(-1)^2(2+2m)=3\cdot1^2(1+m)-2-2m,$$ which gives $$m=-1$$ and $$y=-x+2.$$