Find an equation of a tangent at $C(3,1)$ on $x^2-y^2 = 8$ with an elementary method of analytical geometry. 
Find an equation of a tangent at $C(3,1)$ on $x^2-y^2 = 8$ with an elementary methods of analytical geometry. So with non calculus method! 


The focuses are at $A(4,0)$ and $B(-4,0)$. It is well known that angle bisector of $\angle ACB$ is a tangent at $C$. Now if we reflect $A$ across this tangent we get $D$ which is on $BC$, so we can get $D$ as intersection of line $BC$ with circle centered at $C$ and radius $r=CA$. Then perpendicular bisector of $AD$ is the tangent.  
This is somehow long process to get a tangent. Is there a shorter elementary way to do it?
 A: Compute the anticlockwise inclination $\theta_1$ of $AC$ to the $x$-axis, and do the same with $\theta_2$ and $BC$. The angle of the tangent at $C$ is then $\frac{\theta_1 + \theta_2}{2}$. Find the gradient $m = \tan \left( \frac{\theta_1 + \theta_2}{2}\right)$, and use the point-gradient formula.
A: Here is the fastest way that I know of to get the angle bisector(s) of two (distinct) lines:
Express the two lines as
$$
ax + by = c\\
dx + ey = f
$$
with $a^2 + b^2 = d^2 + e^2$. Then the sum of the two equations is one of the two bisectors, and the difference is the other
In other words, the two bisectors are
$$
(a + d)x + (b + e)y = c + f\\
(a - d)x + (b - e)y = c - f\\
$$
One of the bisectors will be the tangent and the other will be the normal.
A: The tangent line has the equation $$y=mx+n$$ For $$C(3;1)$$ we get $$y=m(x-3)+1$$ So you have to solve $$x^2-(m(x-3)+1)^2=8$$ Solve this equation for $x$ and set the discriminant equal to zero to compute $m$.Since we have a tangent line so we must get only one solution.
A: We write $x=3+h$ and $y=1+k$ and insert in the LHS polynomial of the equation $x^2-y^2-8=0$, getting thus
$$
(3+h)^2-(1+k)^2-8\ .
$$
From this, the degree zero part (the "coefficients part" $3^2-1^2-8$) must vanish, and vanishes, since $(3,1)$ is a point of the curve, and from the rest we isolate only the linear part, which is $6h+2k$. So the equation of the tangent in the frame $(h,k)$ is (obtained by "dedoubling") $6h+2k=0$. (Terms of degree $\ge 2$ from the Taylor expansion have been remoced.) Coming back to the $(x,y)$-frame, the equation is
$$
6(x-3)+2(y-1)=0\ .
$$
Arguably, we are using calculus. If this is rather the case, then please ignore this answer. (The short version of it would have been only the last displayed line, with the comment of "dedoubling" or alternatively "by ignoring higher Taylor monomials in $(x-3)$ and $(y-1)$.)
A: Here is an idea I just came up. We will use a following lemma.
Lemma: Let the tangent and normal at $C$ meet $y$-axsis, at $E$ and $F$. Then $A,B,C,E$ and $F$ are concyclic. 

Proof: Let $\mathcal{C}$ be circle around $\Delta ABC$. Since tangent at $C$ is an angle bisector for $\angle ACB$ it halves arc $AB$ point $F$ must be on $\mathcal{C}$. And since $\angle ECF = 90^{\circ}$ and $D$ is on $y$-axsis, we see that $E$ is also on $\mathcal{C}$.

Now we can make a following construction: 


*

*Write perpendicular bisector for $AC$ (or $BC$) and let it meet $y$-axis at $D$. 

*Write a circle centered at $D$ with a radius $DA$ (or $DB$) and calculate where it cuts $y$-axsis. We get $F$ and $E$. 

*Write an equation of $CF$ and we are done.

A: It can be shown via elementary methods that the tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $(x_0,y_0)$ on the hyperbola has the equation $$\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1$$ (this expresses the pole-polar relationship of a point on the hyperbola and the tangent there) so you can write down the solution to this problem directly.  
Otherwise, computing the angle bisector seems like it would be fairly quick to do. The two angle bisectors have direction vectors $\lVert\overrightarrow{CA}\rVert \overrightarrow{CB}\pm\lVert\overrightarrow{CB}\rVert \overrightarrow{CA}$, from which you can construct equations for the two lines in short order via the point-normal form. You still have to determine which of them is tangent and which is normal, though, so the slope-discriminant method described in another answer might be even faster overall.
