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Suppose that there is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$.

I would like to figure out an equation that describes tangent line to this hyperbola.

How would I be able to do this using calculus? My calculus trials are bring me some gibberish answers.

I did use the calculus below, but what I want is how to derive the general form equation of $y=mx \pm \sqrt{a^2m^2-b^2}$ where $m$ refers to the slope of a tangent line.

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Let $y=mx+c$ be a tangent of $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

Let $(h,k)$ be the point of intersection.

So, $k=mh+c, b^2h^2-a^2k^2=a^2b^2\implies b^2h^2-a^2(mh+c)^2=a^2b^2$

or,$h^2(b^2-a^2m^2)- 2a^2mch -a^2b^2-a^2c^2=0 $

This is a quadratic equation in $h,$ for tangency, the roots need to be same, to make the two points of intersection coincident.

$\implies (- 2a^2mc)^2=4\cdot (b^2-a^2m^2)(-a^2b^2-a^2c^2) $

$a^2m^2c^2=a^2m^2b^2+a^2m^2c^2-b^4-b^2c^2$ cancelling out $a^2$ as $a\ne0$


$0=a^2m^2-b^2-c^2$ cancelling out $b^2$ as $b\ne0$

$\implies c^2=a^2m^2-b^2\implies c=\pm\sqrt{a^2m^2-b^2}$


Alternatively, we can take the central conic to be $Ax^2+By^2=1$

Applying the same method we get, $ABc^2=A+m^2B$

Here $A=\frac 1{a^2},B=-\frac 1{b^2}\implies c^2=a^2m^2-b^2$


Using calculus as André Nicolas has already done, we find the equation of the tangent to be $$\frac {xx_1}{a^2}-\frac{yy_1}{b^2}-1=0$$

Comparing with $mx-y+c=0$ we get, $$\frac{x_1}{ma^2}=\frac{y_1}{b^2}=\frac{-1} c$$

Now $(x_1,y_1)$ lies on the given curve, so $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$, eliminating $x_1,y_1$, we shall get the desired result.

(2b) we know , the parametric equation of the given curve is $x=a\sec t,y=b\tan t$

So, the equation of the tangent becomes $\frac{x\sec t}a-\frac{y\tan t}b-1=0$

Comparing with $mx-y+c=0$ we get, $$\frac{\sec t}{ma}=\frac{\tan t}b=\frac {-1}c$$

So, $\sec t=-\frac{ma}c,\tan t=-\frac b c$

Now use eliminate $t$.

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It is such a nice equation that we might as well differentiate with respect to $x$ immediately. We get $$\frac{2x}{a^2}-\frac{2yy'}{b^2}=0.$$ Solve for $y'$.

Remark: Alternately, we can solve for $y$ in terms of $x$. We get two equations. Then differentiate. A bit messier than the differentiation above, but doable.

The device we used, implicit differentiation, is very useful. In or situation, we can think of the differentiation of $y^2$ with respect to $x$ as using the Product Rule, or the Chain rule. Note that $y^2$ is a product. Its derivative with respect to $x$ is $yy'+y'y=2yy'$.

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