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I'm asked to calculate the tangent to a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at a point $P(x_1,y_1)$ without using derivatives (or limits), and we cannot use geometric transformations, because they 're not linear when applied to non-bounded curves (this is also something I would like an explanation to).

Thanks in advance!

Blue
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Zero Pancakes
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2 Answers2

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Hint.

You can use the equation of the lines thorough $P$ and take the system with the equation of the hyperbola. $$ \begin{cases}y-y_1=m(x-x_1)\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \end{cases} $$

the value of $m$ such that this system has only one solution it the slope of the tangent. (do you see why?)

If $P$ is a point of the hyperbola you find one only value of $m$.

Do you know how to find this value of $m$?

Emilio Novati
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  • Yes, this seems correct! I will try and solve it now and get back to you. – Zero Pancakes Mar 22 '17 at 21:34
  • You can also my answer here: http://math.stackexchange.com/questions/1888090/why-only-2-tangents-can-be-drawn-to-a-hyperbola/1888131#1888131 or here http://math.stackexchange.com/questions/1921627/line-intersects-conic-at-exactly-one-point-implies-the-line-is-tangent-to-conic/1921649#1921649 – Emilio Novati Mar 22 '17 at 21:37
  • $Δ=0$ for one solution, but this is still very perplex..I don't know if there is a faster way around this.. – Zero Pancakes Mar 22 '17 at 21:47
  • The faster way is using derivative. – Emilio Novati Mar 22 '17 at 22:29
  • We get to a point where we have an equation with constants and $m^2$. Why do we have two m's? – Zero Pancakes Mar 22 '17 at 22:38
  • I’d suggest using a different form of equation of a line so that you don’t run into trouble at the vertices, where the tangent is vertical. – amd Mar 22 '17 at 23:12
  • @amd What kind of equation do you mean? – Zero Pancakes Mar 23 '17 at 08:55
  • $\alpha (x-x_1)+\beta (y-y_1)=0$ works for any line, but $y-y_1=m(x-x_1)$ doesn’t handle vertical lines, which you’ll have at $x=\pm a$. – amd Mar 23 '17 at 20:26
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Use an optical property of the hyperbola: a ray thrown from on focus to $(x1, y1)$ reflects as a continuation of the ray thrown from another focus. The bisector of the angle between the rays is normal to hyperbola.

The foci of your hyperbola are $(0, \pm\sqrt{a^2+b^2})$.

user58697
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