A major advantage to writing conic sections in standard form is the ease with which you can apply techniques like Joachimsthal's notation to solve problems such as finding tangents. In fact this is selling Joachimsthal short: you can do a lot more than finding tangents, and if you've not heard of it then it's worth reading this article in Cut-The-Knot and this more technical exposition by Wilson Stothers. This area used to be a major topic in the British A-level syllabus, and this Math SE question suggests it is still taught at high school level in the Netherlands. Even if this is a bag of tricks you're unfamiliar with, the standard form of conics also makes it straightforward to apply implicit differentiation. In fact in current British A-level textbooks, the chapter on implicit differentiation is where you'd come across finding tangents and normals to conic sections. I'll quote you some of the standard results about circles given in a 1980s A-level textbook, which still gives a sense of the "old skool" way students were taught to do things without resorting to calculus: but even by this time, many conics topics had been removed from the syllabus.
- The general equation of a circle is $$x^2 + y^2 + 2gx + 2fy + c = 0$$ The circle has centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.
- The equation of the circle with $AB$ as diameter where $A$ is $(x_1, y_1)$ and $B$ is $(x_2, y_2)$ is given by $$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$
- To find the equation of a circle passing through three points, substitute the coordinates of each point in turn into the general equation, solve the three simultaneous equations for the values of $g$, $f$ and $c$, then substitute these into the general equation.
- The equation of the tangent at $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $$xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$$ Note the relation between the equations: $x^2 \to xx_1$, $y^2 \to yy_1$, $2x \to (x+x_1)$, $2y \to (y+y_1)$.
- The length of the tangent from $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $$\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$$
- The equation of any circle through the intersection of two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ is given by an equation of the form $$(x^2 + y^2 + 2g_1x + 2f_1y + c_1) + \lambda(x^2 + y^2 + 2g_2x + 2f_2y + c_2) = 0$$ where $\lambda$ is a constant that is usually found from the given conditions. If $\lambda = -1$, the equation reduces to a straight line. This is the common chord of the two circles and is known as their radical axis.
The above makes heavy use of the standard form of the equation for a circle, but a lot of that material will generalise to other conics. For example, using the notation at Cut-The-Knot, say we have the general conic
$$ Ax^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H=0 \tag{1}$$
One advantage of this form is we can (as Stothers does) write our equation succinctly in matrix form as
$$ \begin{bmatrix}x & y & 1 \end{bmatrix} \begin{bmatrix}A & B & F \\ B & C & G \\ F & G & H \end{bmatrix} \begin{bmatrix}x \\ y \\ 1 \end{bmatrix} = 0 \tag{2}$$
or simply $\mathbf{x'Mx} = 0$ where $\mathbf{M}$ is a symmetric 3-by-3 matrix. If the coordinates vector $\mathbf{x}$ looks suspiciously like it should have a $z$ but we have for some reason set $z=1$, then welcome to the realm of homogeneous coordinates and projective geometry — this setting unifies the treatment of conics and makes sense of many of their features, such as tangents and asymptotes, and "ideal points" at infinity. This matrix form gives us a test for whether a conic is degenerate: this occurs if and only if $\det \mathbf{M} = 0$. The determinant of the top left 2-by-2 sub-matrix is $\Delta = AC - B^2$ and gives a test to classify conics as ellipses, parabolas or hyperbolas depending on whether $\Delta > 0$ (ellipse, or in the degenerate case, a single point), $\Delta = 0$ (parabola, or if real solutions exist in the degenerate case, two possibly coincident parallel lines) or $\Delta < 0$ (hyperbola, or in the degenerate case, two intersecting lines).
Writing the conic as $(1)$ or $(2)$ lets us generalise the technique for finding the equation of a circle from three points, to find the equation of a general conic if given five points. (There are six coefficients and only five simultaneous equations, but solutions for the coefficients for the general conic can differ by a multiplicative constant. For the standard equation of a circle we fixed the $x^2$ as $y^2$ coefficients to one, so this issue didn't arise, but for a general conic we aren't guaranteed any particular coefficient is non-zero, so can't necessarily pick one to fix in this way.)
The textbook extract showed how to find the pencil of circles passing through the intersection points of two given circles, by taking the weighted sum of the circle's equations (in standard form). The textbook weighted the first equation by one and second by $\lambda$, so $\lambda = -1$ gave the degenerate case of the radical axis, $\lambda = 0$ recovers the first circle, but the second circle is only recovered in the limit $\lambda \to \infty$. To keep things real, you may prefer to weight the first equation by $\lambda$ and the second by $\mu$, not both zero. Then $(\lambda, \mu)=(1,0)$ recovers the first circle, $(0,1)$ recovers the second circle, and $(1,-1)$ gives the radical axis. This doesn't really give us an extra degree of freedom since multiplying $\lambda$ and $\mu$ by a common factor produces the same circle: all that matters is the ratio $\lambda:\mu$ and we are once again in the realm of homogeneous coordinates. (What the textbook denoted by $\lambda$ is in our notation the ratio $\mu/\lambda$, which is often what we want to work with anyway.) We can similarly find a pencil of conics through the intersection points of two given conics $\mathbf{x'Mx} = 0$ and $\mathbf{x'Nx} = 0$ by taking the linear combination $\mathbf{x}'(\lambda \mathbf{M} + \mu \mathbf{N})\mathbf{x} = 0$. We illustrate this for the circle $x^2 + y^2 - 5 = 0$ (yellow) and hyperbola $x^2 - y^2 - 3 = 0$ (blue). The four common points are the base points of the pencil, and for any other point there is only one conic of the pencil that passes through it.
Note that in $\mathbb{R}^2$ conics can intersect in 0, 2 or 4 (possibly coincident) points. If two given circles intersect each other at 2 points, the procedure above for the pencil of circles generates all circles passing through their intersection points, including the degenerate case of the radical axis. If the two points are coincident, the generated circles and radical axis will all be mutually tangential there. The procedure for the pencil of conics will similarly find all conics passing through the intersection points of two given conics, but only if they intersect in 4 (possibly coincident) points. Even if the given conics intersect only in 2 points, the pencil gives a method to find the intersection points of two conics: find (one of) the degenerate conics in the pencil by obtaining a solution to $\det (\lambda \mathbf{M} + \mu \mathbf{N}) = 0$. This involves solving a cubic equation in $\mu/\lambda$ (or just $\lambda$ if you use the A-level textbook's approach and solve $\det (\mathbf{M} + \lambda \mathbf{N}) = 0$ instead). Find the equations of the lines that make up your degenerate conic: if you have a degenerate parabola or hyperbola, you'll need to factorise. Now find the intersection of your lines with one of your given conics. In the example illustrated above, the degenerate cases occur when $\lambda:\mu = -1:1$ (giving the degenerate parabola $2-2y^2-0$ which factorises as $(y+1)(y-1)=0$ i.e. the parallel lines $y=\pm 1$), when $\lambda:\mu = -3:5$ (giving the degenerate hyperbola $2x^2 - 8y^2 = 0$ which factorises as $(x+2y)(x-2y)=0$ i.e. the intersecting lines $y=\pm \frac{1}{2}x$) and when $\lambda:\mu=1:1$ (giving the degenerate parabola $2x^2-8=0$ which factorises as $(x+2)(x-2)=0$ i.e. the parallel lines $x=\pm 2$). Clearly the intersection points are $(2,1)$, $(2,-1)$, $(-2,1)$ and $(-2,-1)$.
Now we introduce the notation of Ferdinand Joachimsthal. Define for points $P(x_i, y_i)$ and $Q(x_j, y_j)$ a quantity $s_{ij} = \mathbf{x}_i' \mathbf{Mx}_j$. This is equivalent to taking the standard equation of the conic and transforming the quadratic terms as $x^2 \to x_i x_j$, $y^2 \to y_i y_j$ and $2xy \to x_i y_j + x_j y_i$, and the linear terms as $2x \to x_i + x_j$ and $2y \to y_i + y_j$:
$$s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H$$
Note the symmetry $s_{ij} = s_{ji}$, and that point $P(x_i,y_i)$ lies on the conic if and only if $s_{ii}=0$. The textbook extract above shows that in the special case of a point outside a circle, and with the $x^2$ and $y^2$ coefficients standardised to one, then $s_{ii}$ is the squared length of the tangent from that point to the circle so naturally becomes zero if we move that point onto the circle. In fact $s_{ii}$ is just the power of a point with respect to the circle, a definition that also extends to the interior of the circle and ties together the intersecting chords theorem, intersecting secants theorem and tangent-secant theorem.
The equation of a general conic can be written as $s=0$ where $s = \mathbf{x'Mx}$ is like $s_{ii}$ but with the specific point $P(x_i,y_i)$ replaced by the general point $(x,y)$, so $s$ is the left-hand sides of equations $(1)$ and $(2)$.
Keeping $P(x_i, y_i)$ as a specific point, but replacing $x_j$ and $y_j$ associated with point $Q$ by a completely general $x$ and $y$, we can also define $s_i = \mathbf{x}_i' \mathbf{Mx}$, i.e.
$$ s_{i} = Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H$$
Then if point $P(x_i, y_i)$ lies on the conic, the equation of the tangent to the conic at $P$ is $s_i = 0$, i.e.
$$ Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H=0$$
The method for finding the tangent to a circle quoted above was just a special case of this.
The condition for the line $PQ$ to be a tangent to the conic is $s_{ij}^2 = s_{ii}s_{jj}$. We can use this condition to find the equations of the tangents to a conic passing through a given point $P(x_i, y_i)$: we replace $Q(x_j,y_j)$ by the general point $(x,y)$ and rewrite the tangency condition as $s_i^2 - s_{ii}s = 0$. The left-hand side is a quadratic in $x$ and $y$. Factorise it into two linear factors: setting each factor to zero gives the equation of a tangent passing through $P(x_i,y_i)$.
If $P(x_i,y_i)$ lies on the tangent to the conic at $Q(x_j,y_j)$, then $$s_{ii} = A(x_i - x_j)^2 + 2B(x_i - x_j)(y_i - y_j) + C(y_i - y_j)^2$$
This generalises the tangent-length formula for a circle quoted above: put $A=C=1$, $B=0$, and the right-hand side is the square of the length $PQ$. The right-hand side is in general equal to $s_{ii} - 2s_{ij} + s_{jj}$: the algebraic proof boils down to using the symmetry of $\mathbf{M}$ and factorising $\mathbf{x}_i' \mathbf{Mx}_i - \mathbf{x}_i' \mathbf{Mx}_j - \mathbf{x}_j' \mathbf{Mx}_i - \mathbf{x}_j' \mathbf{Mx}_j$ as $(\mathbf{x}_i - \mathbf{x}_j)' \mathbf{M} (\mathbf{x}_i - \mathbf{x}_j)$, where $(\mathbf{x}_i - \mathbf{x}_j)'$ is $\begin{bmatrix}x_i - x_j & y_i - y_j & 0 \end{bmatrix}$. Since the third component is zero, only the coefficients $A$, $B$ and $C$ from the top left two-by-two submatrix of $\mathbf{M}$ appear in our final equation. We have $s_{jj} = 0$ since $Q$ lies on the conic, and $s_{ij} = 0$ by the tangency condition, so $s_{ii} - 2s_{ij} + s_{jj}$ reduces to $s_{ii}$ when $PQ$ is a tangent to the conic at $Q$ .
Joachimsthal's notation gives a powerful way to explore poles and polars with respect to a conic. For a given point $P(x_i, y_i)$, there is an associated polar line with respect to the conic, whose equation is simply $s_i = 0$, and we say $P$ is the pole of that line (again, with respect to the given conic). Stothers treats this both algebraically and geometrically. We saw above the special case that when $P$ lies on the conic, its polar line with respect to the conic is the tangent to the conic at $P$. Vice versa, if a line is tangential to the conic then its pole with respect to that conic is the point of tangency. From the symmetry of $s_{ij}=s_{ji}$, it is trivial to prove La Hire's theorem that if $P$ lies on the polar of $Q$, then $Q$ lies on the polar of $P$ (with respect to the same conic). So as a practical example, if a line intersects the conic at two points $P$ and $Q$, and the tangents to the conic at $P$ and $Q$ intersect at $R$, then $R$ is the pole of the original line $PQ$ with respect to the conic. This follows since $R$ lay on both the polar of $P$ and polar of $Q$ so, by La Hire's theorem, $P$ and $Q$ both lie on the polar of $R$, and these two distinct points suffice to define the polar line. Vice versa, given a point through which we can find two tangents to the conic, its polar is the line through the two points of tangency.
There are many other uses for this notation, including finding the common tangents to two conics and determining the minor radius of an ellipse from its major axis and a tangent line, but I'll leave it there. An excellent book for further reading is Geometry by David A. Brannan, Matthew F. Esplen and Jeremy J. Gray, Cambridge University Press (1st edition 2002, 2nd edition 2012).
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