Show that the tangent to the hyperbola $(x_0 , y_0)$ does not intersect the curve anywhere else.

Question

Question: Consider the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Given that the equation of the tangent at the point $(x_0 , y_0)$ is $\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$

Show that the tangent to the hyperbola $(x_0 , y_0)$ does not intersect the curve anywhere else.

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What I have done:

First making y the subject of this:

$$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$$

$$ \Leftrightarrow \frac{xx_0}{a^2} + 1 = \frac{yy_0}{b^2} $$

$$ \Leftrightarrow \frac{xx_0b^2}{a^2} + b^2 = {yy_0} $$

$$ \therefore y=\frac{xx_0b^2}{a^2y_0} + \frac{b^2}{y_0}$$

$$ \Leftrightarrow y=\frac{xx_0b^2}{a^2y_0} + \frac{b^2a^2}{y_0a^2}$$

$$ \Leftrightarrow y= \frac{xx_0b^2+b^2a^2}{a^2y_0} $$

$$ \Longrightarrow y^2 =\frac{x^2x_0^2b^4 + 2xx_0b^4a^2+b^4a^4}{a^4y_0^2} $$

Now

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

$$ \Leftrightarrow \frac{x^2}{a^2} - \frac{\frac{x^2x_0^2b^4 + 2xx_0b^4a^2+b^4a^4}{a^4y_0^2}}{b^2} = 1 $$

$$ \Leftrightarrow \frac{x^2}{a^2} - \frac{x^2x_0^2b^4 + 2xx_0b^4a^2+b^4a^4}{a^4y_0^2b^2} = 1$$

$$ \Leftrightarrow \frac{x^2}{a^2} - \frac{b^2(x^2x_0^2b^2 + 2xx_0b^2a^2+b^2a^4)}{a^4y_0^2b^2} = 1$$

$$ \Leftrightarrow \frac{x^2}{a^2} - \frac{x^2x_0^2b^2 + 2xx_0b^2a^2+b^2a^4}{a^4y_0^2} = 1$$

$$\Leftrightarrow \frac{x^2a^2y_0^2}{a^4y_0^2} - \frac{x^2x_0^2b^2 + 2xx_0b^2a^2+b^2a^4}{a^4y_0^2} = \frac{a^4y_0^2}{a^4y_0^2}$$

$$\Leftrightarrow \frac{x^2a^2y_0^2}{a^4y_0^2} - \frac{x^2x_0^2b^2 + 2xx_0b^2a^2+b^2a^4}{a^4y_0^2} - \frac{a^4y_0^2}{a^4y_0^2} = 0 $$

$$ \Leftrightarrow \frac{x^2a^2y_0^2 -x^2x_0^2b^2 - 2xx_0b^2a^2-b^2a^4 -a^4y_0^2 }{a^4y_0^2} $$

$$ \Leftrightarrow x^2a^2y_0^2 -x^2x_0^2b^2 - 2xx_0b^2a^2-b^2a^4 -a^4y_0^2 = 0 . $$

Now I am stuck now.. How do I proceed? Am I even on the right approach?

Answer 1

hint: Treat the equation you got as a quadratic equation, and show: $\triangle = 0$

Answer 2

The approach in the post is computationally complicated. Almost all lines have equations of the shape $y=mx+c$. (The exception is vertical lines.)

Now consider the equation $\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$. This is a quadratic, so if it has a double root (tangency), then it has no other roots.

For completeness we deal with vertical tangents. From the equation of the tangent line, this can only happen for the tangent line at a point $(\pm a,0)$. The line $x=a$ (or $x=-a$) only meets the hyperbola in one point.

Answer 3

Def'n: $\cosh t=(e^t+e^{-t})/2\;$ and $\;\sinh t=(e^t-e^{-t})/2.\; $ Observe that $\cosh (-t)=\cosh t>0$ and $\sinh (-t)=-\sinh t.\;$ By direct computation we have $$\cosh t \cosh u-\sinh t \sinh u=\cosh (t-u) \quad \text {for all } t,u.$$

Any point on the hyperbola is equal to $(\pm a\cosh t, b\sinh t)$ for some real $t.$ So let $(x_0,y_0)=(\pm a\cosh t_0,b\sinh t_0).$

If $(x,y)$ is on the tangent line and on the hyperbola, let $(x,y)=(\pm a \cosh t,\ b\sinh t)$ with real $t.$

... (i)...If $x<0 \leq x_0$ or if $x_0<0\leq x,$ then $$1=x x_0/a^2- y y_0/b^2=- \cosh t_0 \cosh t - \sinh t_0 \sinh t=-\cosh (t_0+ t) $$ which is impossible because $\cosh r >0$ for all real $r.$

... (ii)... If $x\leq 0\geq x_0$ or if $x\geq 0\leq x_0$ then $$1=x x_0/a^2-y y_0/b^2=\cosh (t_0-t).$$

This implies $t=\ t_0,$ because for real $\;t_0,\;t, $ we have $1=\cosh (t_0- t)\iff$ $ 2 e^{(t_0- t)}=$ $e^{2(t_0-t)}+1\iff$ $ 0=(e^{(t_0- t)}-1)^2 \iff$ $ 1=e^{t_0- t}\iff 0=t_0- t.$

So $(x,y)=(\pm x_0,y)$ but we also have, from (ii) that $x\geq 0\iff x_0\geq 0,$ so $x=x_0$ and $(x,y)=(x_0,y_0).$