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Find an equation for a tangent line to the curve $$ x^2 - y^2 = 5$$ that passes through the point $(1, 1)$.

I realize that I have to use implicit differentiation $$2x - 2y \frac {dy}{dx} = 0$$ $$\frac {dy}{dx} = \frac xy$$

However I do not know how to relate this to the point in order to find the equation of the line that passes through the point, even though the point isn't (presumably) on the curve.

Thank you in advance for the help.

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    $\begingroup$ Hold on... The curve is a circle, and the point given is inside the circle. There is no tangent line that passes through the circle! $\endgroup$ Dec 7 '16 at 9:31
  • $\begingroup$ Messed up, updated it now. It was supposed to be minus not plus. $\endgroup$ Dec 7 '16 at 9:47
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Here's an approach without (implicit) differentation.

A line passing through $(1,1)$ and with slope $m$ has an equation of the form: $$y=m(x-1)+1 \iff y = mx-m+1$$ The points of intersection of this line and the hyperbola $x^2-y^2=5$ are the solutions to the following system: $$\left\{\begin{array}{l} y = mx-m+1 \\ x^2-y^2=5 \end{array}\right.$$ Substitution of the first into the second equation yields a quadratic equation in $x$: $$x^2-(mx-m+1)^2=5 \iff \left( 1-m^2 \right)x^2 + \left(2 m^2- 2 m \right) x - m^2 + 2 m - 6 = 0 $$ The line is tangent to the hyperbola if the discriminant is $0$, so: $$\begin{align}\left(2 m^2- 2 m\right)^2-4\left(1-m^2\right)\left(- m^2 + 2 m - 6 \right) = 0 & \iff -16 m^2 - 8 m + 24 = 0 \\ & \iff 2m^2 + m -3 = 0 \\ & \iff \left( m-1 \right)\left(2m+3 \right) = 0 \end{align}$$ This gives you two slopes.

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  • $\begingroup$ This is a really cool solution. And it will help me with visualizing the problem in the future. Thanks! $\endgroup$ Dec 9 '16 at 22:43
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Continuing from where you left off, we have:

$$x^2-y^2=5$$

$$\frac{\text dy}{\text dx}=\frac x y$$

A line that passes through points $(1, 1)$ and some point $(a, b)$ must have the form:

$$y-1=\frac{b-1}{a-1}(x-1)$$

If $(a,b)$ lies on the hyperbola then $a^2-b^2=5$.

If the line is tangent to the hyperbola then:

\begin{align} \frac{b-1}{a-1} &= \frac a b\\\\ a-b &= a^2-b^2\\ &=5 \end{align}

This means that:

\begin{align} (5+b)^2-b^2&=5\\ 25+10b&=5\\\\ b&=-2\\ a&=3 \end{align}

Thus the equation of the tangent that passes through $(1,1)$ is:

$$y-1=\frac{-3}{2}(x-1)$$

StackTD provides an interesting solution, however notice that the one of the solutions given, namely $m=1$, results in an asymptote to the hyperbola. One might be a little pressed to call this a tangent...

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  • $\begingroup$ Well there is no problem if you take into account a point at infinity, or just plainly switch to projective space. $\endgroup$
    – Ennar
    Dec 7 '16 at 11:38
  • $\begingroup$ That's true! Consider me a little pressed. $\endgroup$ Dec 7 '16 at 11:58
  • $\begingroup$ Thank you! This is the type of solution I was looking for. $\endgroup$ Dec 9 '16 at 22:41
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The equation of tangent at $(\sqrt 5\sec2u,\sqrt5\tan2u)$ will be

$$x(\sqrt5\sec2u)-y(\sqrt5\tan2u)=5\iff x-y\sin2u=\sqrt5\cos2u$$

As it has to through $(1,1);$ $$1-\sin2u=\sqrt5\cos2u$$

Now use Weierstrass substitution to find either $$\tan u=1\implies\cos2u=?,\sin2u=?$$

or $$\tan u=\dfrac{\sqrt5-3}2$$

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