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In my Calculus book it says:

We have shown that planes are represented by first-degree equations and cones by second-degree equations. Therefore, all conics can be represented analytically (in terms of Cartessian coordinates $x$ and $y$ in the plane of the conic) by a second-degree equation of the general form $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,$$ where $A, B, ..., F$ are constants...

How can the conclusion about the general form of the equation of a conic be drawn from the premises in the first sentence?

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  • $\begingroup$ Apply a "rigid" transformation that moves the "plane of the conic" to the $xy$-plane itself. The transformed cone will still have a second-degree Cartesian equation in $x$, $y$, $z$; we get the equation of cone's intersection with the now-conveniently-placed plane of the conic by setting $z=0$, leaving the general $xy$-form shown. $\endgroup$
    – Blue
    Commented Jul 5, 2019 at 11:47

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In fact, the intersection of a plane with any quadric surface results in a conic. I find it most convenient to derive this using the homogeneous matrix form of these equations.

Letting $\mathbf X = (X,Y,Z,1)^T$, consider the quadric surface with the implicit equation $\mathbf X^TQ\mathbf X=0$, where $Q$ is a symmetric $4\times4$ matrix. If we write $\mathbf x=(x,y,1)^T$, where $x$ and $y$ are the coordinates of a point on the plane in its coordinate system, then there is a $4\times3$ matrix $M$ such that $\mathbf X=M\mathbf x$ is the mapping from plane coordinates to space coordinates. In particular, if $\tilde{\mathbf X}$ and $\tilde{\mathbf Y}$ are the (inhomogeneous) coordinates in $\mathbb R^3$ of the unit coordinate vectors of the plane’s coordinate system, and $\tilde{\mathbf C}$ the origin of this coordinate system, then $$M = \begin{bmatrix}\tilde{\mathbf X} & \tilde{\mathbf Y}&\tilde{\mathbf C}\\0&0&1\end{bmatrix}$$ is one such matrix. (In fact, every parameterization of the plane has a corresponding matrix $M$: $\mathbf X=M\mathbf x$ is just a compact way to write this parameterization.)

Substituting into the equation of the quadric, we have $$\mathbf X^TQ\mathbf X = (M\mathbf x)^TQ(M\mathbf x) = \mathbf x^T(M^TQM)\mathbf x = 0. \tag{*}$$ Now, $M^TQM$ is itself a symmetric matrix, so if $$M^TQM=\begin{bmatrix} A & \frac B2 & \frac D2 \\ \frac B2 & C & \frac E2 \\ \frac D2 & \frac E2 & F\end{bmatrix}$$ then equation (*) expands into $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. So, every intersection of a plane with this surface has an equation of this form in the plane’s Cartesian coordinate system.

Every circular cone is an affine image of the cone $x^2+y^2=z^2$, so its implicit equation is, in homogeneous matrix form, $\mathbf X^T(A^T\operatorname{diag}(1,1,-1,0)\,A)\mathbf X=0$, where $A$ is the matrix of an invertible affine transformation. The central matrix in this equation is clearly symmetric, so this is a special case of the above and therefore every conic has a general equation of the required form. Actually, to cover every type of conic, one also has to consider intersections of a plane with a cylinder, which can be considered a degenerate cone, but clearly that’s also a special case of the above derivation.

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Well, it's a bit of a stretch. Take a conic, represented by a homogeneous quadratic like $$ ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0, $$ which is what I assume was meant by "cones are represented by second-degree equations." Then take a plane, represented by something like $$ px + qy + rz + s = 0. $$ Supposing, for a moment, that $r \ne 0$, we can write an equivalent representation, namely $$ \frac{p}{r}x + \frac{q}{r}y + z + \frac{s}{r} = 0, $$ which I'll rewrite by assigning new names, as $$ p'x + q'y + z + s' = 0 $$ and then find, that for every point $(x, y, z)$ on this plane, we have $$ z = -s' -p'x -q'y $$ So if a point lies on this plane, it looks like $$ (x, y, z) = (x, y, -s' -p'x -q'y) $$ Furthermore, the numbers $x$ and $y$ constitute 'coordinates' for that plane, i.e., every point of the plane corresponds to a unique $xy$-pair, and vice versa.

Now if that point also lies on the conic I mentioned earlier, then we must have $$ ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0, $$ which means (replacing $z$ with $-s' -p'x -q'y$) that $$ ax^2 + by^2 + c(-s' -p'x -q'y)^2 + dxy + ex(-s' -p'x -q'y) + fy(-s' -p'x -q'y) = 0, $$ which we can multiply out, and gather like terms in powers of $x$ and $y$ to get something that looks like $$ (a + p'^2 -ep')x^2 + (b + q'^2-fq')y^2 + dots = 0. $$ Letting $A = (a + p'^2 -ep'), B = (b + q'^2-fq'), \ldots,$ we have exactly the form promised in the quoted section of your question.

Of course, this all assumed that $r \ne 0$. In the case where $r = 0$, we know that one of $p$ and $q$ must be nonzero, and we simply do the same thing with whichever one is nonzero (or pick one at random, if both are nonzero). Suppose that $q \ne 0$. Then we end up with an expression for $y$ in terms of $x$ and $z$, and we discover that $x$ and $z$ work as coordinates for the plane, and we end up with a quadratic in $x$ and $z$ instead of in $x$ and $y$, but that's still fine. Similarly, if $p \ne 0$, we can end up with a quadratic in $y$ and $z$.

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