I am working through some calculus problems (this is in a section on implicit differentiation) and this one is giving me trouble.

I am trying to find the equations of the tangent lines to the hyperbola $$xy=1$$ That pass through the point $(-1,1)$.

As in the other problems of this type, I implicitly differentiated the relation between $x$ and $y$ $$\frac{dy}{dx}x+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$$ Then we have: $$\frac{dy}{dx}\vert_{(-1,1)}=-(1/-1)=1$$ So I would think lines passing through this point would have the equation: $$y-1=x+1\Rightarrow y=x+2$$ But this is definitely not true from plotting on wolfram alpha. The solutions in the book are $$y=\pm(2\sqrt{2}-3)x-2\pm2\sqrt{2} $$

  • $\begingroup$ Tangent at what point of the hyperbola? $\endgroup$
    – user258700
    Jun 21 '16 at 0:53
  • $\begingroup$ @AhmedHussein sorry, editing. $\endgroup$ Jun 21 '16 at 0:53
  • $\begingroup$ @ahmedhussein the OP says $(-1,1)$, but something tells me this is a typo because the point is not on the curve. $\endgroup$
    – zz20s
    Jun 21 '16 at 0:54
  • $\begingroup$ @zz20s if so, it is a typo in the book and i feel absolved. $\endgroup$ Jun 21 '16 at 0:54
  • 1
    $\begingroup$ @qbert Hm... I'd guess it's a typo, since $-1\cdot 1 \neq 1$. $\endgroup$
    – zz20s
    Jun 21 '16 at 0:55

They want the tangent line(s) that pass through the point $(-1,1)$, which is clearly not on the curve.

Let the point of tangency be $(a,b)$. Then the slope of the tangent line, by your calculation, is $-b/a$. The equation of the tangent line is $$y-b=(-b/a)(x-a).$$ Since this line passes through $(-1,1)$, we have $$1-b=(-b/a)(-1-a).$$ Using the fact that $b=1/a$ we get $$1-\frac{1}{a}=(-1/a^2)(-1-a)$$ This simplifies to $a^2-2a-1=0$. Now we can solve for $a$ and finish.


Using homogenous coordinates, your hyperbola can be expressed as the solution set to $P(X) = X^TQX = X^T\begin{pmatrix}0&1&0\\1&0&0\\0&0&-2\end{pmatrix}X = 0$

What makes this form neat is that the tangent line at any point of this parabola can be calculated simply by

$QX \equiv Q\begin{pmatrix}x\\y\\1\end{pmatrix}$ where $x,y$ denote the coordinates in the Euclidean plane.

Since the line must also pass through (-1,1) we need only solve

$\begin{pmatrix}-1&1&1\end{pmatrix}Q\begin{pmatrix}x\\y\\1\end{pmatrix}=0\\ \begin{pmatrix}-1&1&1\end{pmatrix}\begin{pmatrix}y\\x\\-2\end{pmatrix}=0\\$

which gives $x-y-2=0$ We also have the condition $y = 1/x$, combined we have $x^2-2x-1=0$ which can easily be solved.


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