Equation of a tangent line to a lemniscate

Question

Find the equation of the tangent line to $$(x^2 + y^2)^3 = x^2 - y^2$$ at the point $(0, 0)$.

This is the problem I'm encountering: after taking the implicit derivative, I plug $(0, 0)$ in. Everything cancels out and I get the equation $0 = 0$.

Answer 1

This is a figure 8 which crosses the origin and has two possible tangents.

From $(x^2 + y^2)^3 = x^2 - y^2 $, I get

$\begin{array}\\ 2x-2yy' &=3(2x+2yy')(x^2+y^2)^2\\ &=3(2x(x^2+y^2)^2+2yy'(x^2+y^2)^2)\\ \text{or}\\ 2yy'(3(x^2+y^2)^2+1) &=2x(1-3(x^2+y^2)^2)\\ \text{or}\\ yy'(3(x^2+y^2)^2+1) &=x(1-3(x^2+y^2)^2)\\ \end{array} $

At the origin, and this is the key observation, $y \approx xy' $ so this becomes $xy'^2(3(x^2+y^2)^2+1) \approx x(1-3(x^2+y^2)^2) $ or $y'^2(3(x^2+y^2)^2+1) \approx (1-3(x^2+y^2)^2) $.

Setting $x = y = 0$, $y'^2 \approx 1 $ so $y' = \pm 1$.

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Here is a simpler way (which I thought of after submitting the above).

Let $y = ax$ at the origin.

From $(x^2 + y^2)^3 = x^2 - y^2 $ we get $(x^2 + a^2x^2)^3 = x^2 - a^2x^2 $ or $x^6(1 + a^2)^3 = x^2(1 - a^2) $ or $x^4(1 + a^2)^3 = (1 - a^2) $.

Setting $x=0$, $1-a^2 = 0$ or $a = \pm 1$.

Answer 2

Alternatively, in polar form, the curve is $$r^6=r^2\cos2\theta$$ and $r\neq0$

As $r\rightarrow 0$, $\cos 2\theta\rightarrow 0$ so the tangents at the origin are $\theta=\pm\frac{\pi}{4}$ i.e. $$y=\pm x$$