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Is there a way to show this other than with calculus and implicit differentiation? I am able to easily do this, but I was wondering if there was another way to show it. I feel I may be missing something obvious or going down unnecessary rabbit trails because I have tried using the Diophantus chord method.

Using the chord method, I have that the equation of the tangent line would have the form $Y=t(X-3)+5$ where t is some rational slope.

I then substituted this into $0=X^3-Y^2-2$.
Eventually, I get to $0=(X-3)(X^2+(3-t^2)X+(9+3t^2))$.

The first factor gives the known solution of $X=3$ and I tried to use the quadratic formula to solve the second factor but then only led me a nasty looking expression with a square root, which leads me to believe I am wrong since I am looking only for rational solutions. I also feel that I have gone down the wrong path because even if I do have success with the chord method, I would only have parameterized the rational solutions to $y^2=x^3-2$ and not actually found the equation of the tangent line.

If someone could guide me toward a method of finding the tangent to $y^2=x^3-2$ at (3,5) without the use of calculus (or point out what I am missing) that would be awesome. Thank you!

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Y^2 = x^3 + 2

2y. dy/dx = 3x^2

The gradient of the tangent to the curve at (3,5) is

  1. dy/dx|(x=3 ,y=5) = 27

m= 27/10

since (3,5) lies on the curve y - 5 = 27/10(x-3)

y= (-81+50)/10 + 27x/10

y = 27x/10 - 31/10 is the required equation of the tangent the that curve at the rational point (3,5)

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  • $\begingroup$ Sketch the graph of the function, if you don't want calculus $\endgroup$ – Abdulhafeez Abdulsalam Feb 17 '18 at 22:32
  • $\begingroup$ I don't understand why this has been upvoted. OP explicitly asked for a solution that doesn't use calculus. $\endgroup$ – pjs36 Feb 17 '18 at 23:20
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Another method, borrowed from algebraic geometry:

Change origin, setting $x=3+h$, $y=5+k$. The equation of the tangent is the linear part of the equation in $h,k$. Thus \begin{align}y^2=x^3-2&\iff (5+k)^2=(3+h)^3-2\iff 25+10k+k^2=27+27h+9h^2+h^3-2\\ &\iff h^3+9h^2-k^2+27h-10k=0. \end{align} So the equation of the tangent w.r.t. the new origin is $$27h-10k=0\iff27(x-3)-10(y-5)=0\iff 27x-10y=31.$$

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  • $\begingroup$ Thanks for the answer, that is a slick way to do it! I have one question, how do we know that the linear part of the h,k equation is going to be the equation of the tangent at x=0? $\endgroup$ – Aaron Hultstrand Feb 18 '18 at 15:36
  • $\begingroup$ This is a polynomial function (in two variables), and so it is equal to its Taylor expansion. In a Taylor expansion, the linear part corresponds to the tangent. $\endgroup$ – Bernard Feb 18 '18 at 16:09

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