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Suppose we have a function $y^2 = x^3 + Ax + B$ which we differentiate implicit to find $$\frac {dy} {dx} = \frac {3x^2 + A} {2y}$$

Now suppose we know a point $(x_1,y_1)$ on the curve. Define $$y = m(x - x_1) + y_1$$

Write $$(m(x - x_1) + y_1)^2 = x^3 + Ax + B$$ Why is $x_1$ then a double root of this equation ?

The question arises in Eliptic Curve theory. I see that our line is tangent to the curve, but no more.

Is it always true that if a line $L$ defined by $y = m(x-x_1) + y_1$ is tangent at $x_1$ to the curve $y = f(x)$ then for $ m(x-x_1) + y_1 = f(x)$ we have $x_1$ a double root ? Should $y$ be squared and does it also hold for $y^k$ where $k > 2$ ?

Shuzheng
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    Your $f$ needs to be (locally) analytic in terms of $x$. If you have a more complicated curve, say given by a polynomial equation $G(x,y)=0$, and if the partials are not both zero, then looking at the linear equation $y=mx+b$ passing through a point of the curve and with the right slope, then $G(mx+b,x)$ should be a polynomial in $x$ with (at least) a double root at the $x_0$ that belonged to the point in question. The $y$ is squared here ’cause it appears to power $2$ in the original. – Lubin Nov 28 '14 at 19:38

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You are asking why $f=x^2 - 2x_1x+x_1^2$ divides the difference of the right and left of your equation. Equivalently, why $\,f$ divides $2my_1(x-x_1) +y_1^2-x^3-Ax-B$. But $2my_1=3x_1^2+A$, and $y_1^2 = x_1^3+Ax_1+B$. I think you can do the rest.

Lubin
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  • So it holds for $ k> 2$ ? Is it a double root by some geometric property or is it arbitrary that it happens to be a double root ? – Shuzheng Nov 30 '14 at 16:02
  • You should expect the tangent line to have at least a double contact with the given curve. With pairs of ellipses, you can get double, triple, and quadruple contacts, just by arranging positions right. – Lubin Nov 30 '14 at 18:25
  • But the tangent line only intersects at one point, so why then a double root ? – Shuzheng Nov 30 '14 at 18:41
  • Have you looked at examples? What about the simple example of $y=x^2-2x+1$ and the $x$-axis? The parabola touches the axis at only the point $(1,0)$, but when you eliminate the $y$ from the two equations $y=0$ and the $y=(x-1)^2$, you get the equation $0=(x-1)^2$. Please do some computations. – Lubin Nov 30 '14 at 18:47
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    Sorry, if I've given you the expression that I don't do computations. What I mean is this: suppose we have a function $f(x)$ and then define the line $y = f^{'}(x_1)(x - x_1) + y_1$. Is it then true that (for any function $f(x)) that $f^{'}(x_1)(x - x_1) + y_1 = f(x)$ has $x_1$ as a double root ? I see this is not the case if $f(x)$ is linear ? So it is not a geometric property that always hold ? I see that the secant line has two roots with $f(x)$. Therefore one should expect that the tangent line has two roots (the same root with multiplicity 2), but this is not always the case, why ? – Shuzheng Nov 30 '14 at 19:37
  • Sorry, if I've given you the expression that I don't do computations. What I mean is this: suppose we have a function $f(x)$ and then define the line $y=f^{′}(x_1 )(x−x_1 )+y_1$. Is it then true that (for any function $f(x)$) that $f^{'}(x_1)(x - x_1) + y_1 = f(x)$ has $x_1$ as a double root?I see this is not the case if $f(x)$ is linear?So it is not a geometric property that always hold ?I see that the secant line has two roots with $f(x)$. Therefore one should expect that the tangent line has two roots (the same root with multiplicity $2$), but this is not always the case, why ? – Shuzheng Nov 30 '14 at 19:43
  • @user111854: I say old chap, it seems like you've posted the same comment twice, but one has formatting errors making it illegible; yet is seems the content is the same. Would you mind deleting the "bad" one? Regards, and of course, Cheers! – Robert Lewis Nov 30 '14 at 21:11
  • When the tangent line has a triple contact with the curve in question, the nearby secant lines have three intersections with the curve. Like $y=x^3$. – Lubin Nov 30 '14 at 23:12
  • Ahh, I see. But this implication doesn't hold the other way around. – Shuzheng Dec 01 '14 at 07:31
  • Like for $(x-3)(x-5) $. – Shuzheng Dec 01 '14 at 07:33
  • Sorry, don’t understand your last two comments. Let’s continue this by e-mail, shall we? – Lubin Dec 01 '14 at 16:52