(1a)
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 $a\ne0$
$0=a^2m^2b^2-b^4-b^2c^2$
$0=a^2m^2-b^2-c^2$ cancelling out $b^2$ as $b\ne0$
$\implies c^2=a^2m^2-b^2\implies c=\pm\sqrt{a^2m^2-b^2}$
(1b)
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 $A=\frac 1{a^2},B=-\frac 1{b^2}\implies c^2=a^2m^2-b^2$
(2a)
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 $t$.
I will not explain calculus. There are many websites that do it very well, probably much better than I can, so I'll leave it to them. I will address prerequisites for calculus with an emphasis on procedure rather than deep understanding. At some point, you would do well to revisit these concepts in a slow, measured pace, being very careful and filling in all the details. However, your position seems more of an "in over my head" thing so I'll try to address it from that perspective.
There are only two big concepts in calculus, the derivitive (verb: to differentiate) and the integral (verb: to integrate). Both of these are special uses of the extremely broad concept of a limit, but many introductory calculus classes only touch on limits in a very superficial way. Certainly to do a first physics course with calculus you will not need a deep understanding of limits, only an appreciation for why they let us get the results we want.
If you do not have a good grasp on trigonometry, you can still do/understand calculus but it will seem rather artificial. You should know the sine and cosine functions and it will help to know the other four that usually accompany them. You should know the unit circle and the special values on it. Some basic identities will make your life easier.
For limits, you can get by with an intuitive understanding of functions, and of real numbers. It is more helpful to have a good intuition for rational numbers, for example you should know that there are infinitely many rationals in between any two real numbers. You must understand the notion of the domain of a function. Experience with rational functions is extremely valuable. To do calculations with limits, you should be very comfortable with simplifying rational expressions, including domain issues and extraneous solutions. Again, you can do without rational functions, but you are much better equipped to understand the significance of limits if you can manipulate them without much trouble.
For derivatives, you will need to be familiar with operations of functions: addition, subtraction, multiplication, division. A special emphasis on composition of functions: for you this will probably be the most important prerequisite for solving physics problems. You should be familiar with but do not need to be extremely comfortable with implicitly defined curves: for example, the circle is not given in $y=f(x)$ form, but it is still well-defined. You should understand the domain issues that can arise when converting implicitly defined curves into function form.
For integrals, there are a lot of skills you could need, but I will try and keep it to a bare minimum. I would not try to understand the real definition of an integral (your resources may call it a Riemann integral), but it is absolutely essential that you understand the intuition behind it, and its link to limits.
Of course you must be familiar with finding area of basic shapes. You must be extremely comfortable with reading $\Sigma$ notation; if not in reading it directly, then at least to translating it into $+$ notation (you do not need to be able to write $\Sigma$ notation well). However, the most useful skills for cracking integrals are pattern-recognition and persistence; they can sometimes require quite a bit of creativity to solve.
There is another (shorter) list which I think is equally important for your situation: things which you are not expected to know, but are expected to pick up during a calculus class. These include deeply understanding inverse functions, familiarity with theorems of the form "If … then there exists …", high comfort with implicit curves, high comfort with recognizing compositions of functions [you will need to pick this one up], the significance of variables as distinct from numbers, skill at visualizing 3D space, distinction between functions and their values at points.
At some point while learning derivatives, you will come across the notion of related rates. Please learn this very carefully. Many students struggle a lot with this section — including me — but it is a very important use of calculus for physics. Perhaps it will not come up directly in your class, but if you know it well you will see it hiding just behind the things you discuss.
Best Answer
There is no division by zero, because you're not evaluating the $\dfrac{f(a+h)-f(a)}{h}$ at zero. Instead, you're taking a limit, i.e. you're trying to figure out what this expression approaches, as $h$ approaches zero.
There are some times when this limit is not defined. For example, let $f(x)=\sqrt[3]{x}$ and let $a=0$. Then
$$\dfrac{f(a+h)-f(a)}{h}=\dfrac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}=\dfrac{\sqrt[3]{h}}{h}=\dfrac{1}{\sqrt[3]{h^2}}.$$
And this expression will approach $\infty$ as $h$ approaches $0$.