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I working on a question which I can't figure out by myself and I would like your help. It goes as follows.

Show that the tangent to a hyperbola at any point $(x_0,y_0)$ can be described by the equation $$b^{2} x_{0} x-a^{2} y_{0} y=a^{2} b^{2}$$ To do this use the general formula for a hyperbola given by: $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$

What I have already tried

  • My first thougt proces was trying to differentiate the hyperbola function implicitly to $x$, which gives you $\frac{dy}{dx}=\frac{b^2x_0}{a^2y_0}$ (with the point already filled in). Then I used the formula for a tangent line and solved, but althoug it looked similar it wasn't quite the same as mentioned above. I've got $$a^2y^2+a^2y_0y =b^2x_0x-b^2x_0^2$$
  • Then I thougt I maybe didn't understand what was being asked, so maybe the point is not bounded to the curve but could be any point on the $x$ and $y$ axis. So then I solved the hyperbola function for $y$, which gave me a positive and negative solution in terms of $x$. And then I stopped because I didn't think this would me the best way.

so if anyone could help me, please do. It would be very much appreciated ;)

MathFail
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TheCreator
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  • See, for instance, Cut-the-Knot's Joachimsthal's Notations page for the general conic; this answer of mine gives a proof. See also the question "The Instant Tangent". – Blue Jul 30 '22 at 09:01
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    You're nearly there: you dropped a subscript and a sign, but you already have from $ \ y - y_0 \ = \ y'·(x - x_0) \ \ , $ $$ a^2y_0y \ - \ a^2y_0^2 \ \ = \ \ b^2x_0x \ - \ b^2x_0^2 \ \ \Rightarrow \ \ b^2x_0^2 \ - \ a^2y_0^2 \ \ = \ \ b^2x_0x \ - \ a^2yy_0 \ \ . $$ The left side is equal to $ \ a^2b^2 \ \ , \ $ since the equation of the hyperbola can be written as $ \ b^2x^2 \ - \ a^2y^2 \ = \ a^2b^2 \ \ $ and applies to any point on the curve. The equation sought follows. –  Jul 30 '22 at 09:26

1 Answers1

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$$a^2y^2+a^2y_0y =b^2x_0x-b^2x_0^2\tag{1}$$

Your Eq.$(1)$ is incorrect.

Your slope is correct, $k=\frac{dy}{dx}|_{(x_0,y_0)}=\frac{b^2x_0}{a^2y_0}$

So the tangent line equation is:

$$y-y_0=\frac{b^2x_0}{a^2y_0}(x-x_0)$$

Simplify and we get:

$$a^2y_0y-a^2y_0^2=b^2x_0x-b^2x_0^2$$

Further:

$$b^2x_0x-a^2y_0y=b^2x_0^2-a^2y_0^2\tag{*}$$

Because $(x_0,y_0)$ is on the hyperbola, we have $$\frac{x_0^{2}}{a^{2}}-\frac{y_0^{2}}{b^{2}}=1~~\Longrightarrow~~ b^2x_0^2-a^2y_0^2=a^2b^2$$

Plug into Eq.$(*)$

$$b^2x_0x-a^2y_0y=a^2b^2$$

We are done.

MathFail
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