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
$$y=\pm(2\sqrt{2}-3)x-2\pm2\sqrt{2} $$
Best Answer
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.