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$$x(x^2 + y^2) = a(x^2 - y^2)$$

I was trying to find the equation of the tangent as $$(Y-0) = \frac{dy}{dx}(X - 0)$$ where $$\frac{dy}{dx} = \frac{2ax-3x^2-y^2}{2(x+a)y}$$ So here putting the values of $(x,y)$ as $(0,0)$ makes the derivative non existent. I am stuck at this point and unable to proceed further.

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    $\begingroup$ Welcome to Maths SX! This is an algebraic curve passing through the origin. It is known that the set of tangents to an algebraic curve at the origin is given by the homogeneous part of lowest total degree in the equation, – namely $x^2-y^2=0$ if $a\ne 0$. $\endgroup$
    – Bernard
    Sep 19 at 17:33
  • $\begingroup$ @Bernard I think you should write that as an answer. $\endgroup$
    – Math Lover
    Sep 19 at 17:49
  • $\begingroup$ @MathLover: OK, I'll do it if you think it can be useful to other users of this site, mylord (albeit it only answers the question in the title). $\endgroup$
    – Bernard
    Sep 19 at 18:11
  • $\begingroup$ @Bernard at least I found it useful :) I think I had read it at some point in time but I surely did not remember it. $\endgroup$
    – Math Lover
    Sep 19 at 18:14
  • $\begingroup$ I get $2(a+x)y$ in the denominator of $\frac{\mathrm{d}y}{\mathrm{d}x}$. $\endgroup$ Sep 20 at 2:05
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The equation can be written as $$x(x^2 + y^2)- a(x^2 - y^2)=0.$$ It is the equation of an algebraic curve passing through the origin. It is known that the set of tangents to an algebraic curve at the origin is given by the homogeneous part of lowest total degree in the equation. Therefore, this curve has $2$ branches through the origin if $a\ne 0$, and the (global) equation of the tangents to each of these branches at the origin is $$x^2-y^2=(x-y)(x+y)=0. $$

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  • $\begingroup$ nicely written! $\endgroup$
    – Math Lover
    Sep 19 at 18:24
  • $\begingroup$ Thank you for your kind appreciation! $\endgroup$
    – Bernard
    Sep 19 at 18:25
  • $\begingroup$ Thanks. This was more interesting to know. If you'd be kind enough to suggest some book/website where I could learn more about this I'd be grateful. $\endgroup$ Sep 22 at 13:25
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Alternatively, when expressed in polar form, the curve is

$$r\cos\theta=a\cos2\theta$$

The tangent(s) at the pole are given by values of $\theta$ as $r\rightarrow0$

Therefore $\theta=\pm\frac{\pi}{4}\implies y=\pm x$

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  • $\begingroup$ Quite an ingenious approach! (+1) $\endgroup$
    – Bernard
    Sep 19 at 18:28
  • $\begingroup$ Thanks @Bernard $\endgroup$ Sep 19 at 19:43
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(This is essentially the same as Bernard's answer, but hopefully provides some intuition.)

First, recall the standard "little-$o$" notation: if we are interested in limits around $L$, then writing "$o(f(x))$" means "an omitted term $t(x)$ such that $\lim_{x\to L}{\frac{t(x)}{f(x)}}=0$." (Or, less precisely, "an omitted term much smaller than $f(x)$".) In our case, we'll be interested in limits around $0$, so $L=0$.

Now note that the definition of the derivative (or equivalently, Taylor's theorem) gives us an equivalent characterization of tangent lines:

The tangent line at the origin is $y=sx$ iff the original curve is $y=sx+o(x)$.

Tentatively assuming that a tangent line exists for some value of $s$, we can substitute this expression in for $y$ to obtain $$x(x^2+s^2x^2+o(x^2))=a(x^2-s^2x^2+o(x^2))$$ Expanding, $$(1+s^2)x^3+o(x^3)=a(1-s^2)x^2+o(x^2)$$ Dividing by $x^2$, $$(1+s^2)x+o(x)=a(1-s^2)+o(1)$$ Taking the limit as $x\to0$, any term that is $o(1)$, $\sim x$, or $o(x)$ vanishes. Thus $$0+0=a(1-s^2)+0$$ and we can see that the tangency slopes are $s=\pm1$.

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