coordinate systemseuclidean-geometrygeometryplane-geometry
I found this problem in a plane geometry book
Let $P=(x_1,y_1)$ be a point on the curve $Ax^2+By^2+2Hxy+2Gx+2Fy+C=0$ Show that the tangent line at $P$ has equation $Axx_1+Byy_1+H(xy_1+yx_1)+G(x+x_1)+F(y+y_1)+C=0$
I think it might be useful to use the definition of derivative but I'm not completely sure how to proceed? Any suggestion?
While geodude's method works, I'm putting up my calculus version of the problem.
Taking the random curve: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Differentiating it wrt x: $$2ax + 2hx\frac{dy}{dx} + 2hy + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$ $$\implies (ax + hy + g) + \frac{dy}{dx}(hx + by + f) = 0$$ $$\implies \frac{dy}{dx} = -\frac{(ax + hy + g)}{(hx + by + f)}$$
The equation of the tangent at the point $(x_1, y_1)$ is: $(y-y_1) = \frac{dy}{dx} (x - x_1)$
$$\implies (y-y_1)(hx+by+f) = (x_1 - x)(ax + hy + g)$$ $$\implies hxy + by^2 + fy -hxy_1 - byy_1 - fy_1 = axx_1 + hyx_1 + gx_1 - ax^2 - hxy - gx$$
Rearranging: $$ax^2 + by^2 + 2hxy + gx + fy = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1$$
Adding $gx + fy + c$ on both sides: $$ax^2 + by^2 + 2hxy + 2gx + 2fy +c = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1 +gx + fy + c$$
The LHS is $0$.
$$\therefore axx_1 + 2h\bigg(\frac{x_1y + xy_1}{2}\bigg) + byy_1 + 2g\bigg(\frac{x_1 + x}{2}\bigg) + 2f\bigg(\frac{y + y_1}{2}\bigg) + c = 0$$
Which is the "T" form of the equation.
Here is a graphical representation in the case of an ellipse :
*Fig. 1.$
Being given the equation of the conic
$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c\tag{*}$$
Let $U=\pmatrix{u\\v}$ be a directing vector of the chord, allowing its parameterization under the form :
$$\begin{cases}x&=&x_1+pu\\y&=&y_1+pv\end{cases}\tag{**}$$
The coordinates $(x,y)$ of the intersection points $P_1$ and $P_2$ of the chord with the conic curve will verify the expression obtained by "plugging" the expressions of $x$ and $y$ given by (**) into (*) giving rise to a quadratic polynomial in variable $p$:
$$Ap^2+2Bp+C=0 \ \text{with} $$ $$\begin{cases}A&=&au^2 + 2huv + bv^2\\B&=& u(ax_1+hy_1+g)+v(hx_1+by_1+f)\\C&=&ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c\end{cases}\tag{1}$$
Observe now that point $P_1$ is the midpoint of the chord iff the intersection points $Q$ and $Q"$ correspond to two opposite values of $p,-p$. It amounts to say that the sum of the roots of quadratic equation (1) is equal to zero, otherwise said that the central coefficient $2B$ is zero.
Relationship $B=0$ considered as a dot product equal to $0$ expresses the geometrical fact that :
$$U=\pmatrix{u\\v} \ \text{is orthog. to} \ N:=\pmatrix{ ax_1+hy_1+g\\hx_1+by_1+f}\tag{2}$$
$N$ being a normal vector to the chord, the equation of this chord is :
$$(x-x_1)(ax_1+hy_1+g)+(y-y_1)(hx_1+by_1+f)=0\tag{3}$$
Besides, equation $T=S_1$, when expanded, is :
$$axx_1 + h(xy_1+x_1y) + byy_1 + g(x+x_1) + f(y+y_1) + c=ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c\tag{4}$$
which can be put under the form :
$$x(ax_1+hy_1+g)+y(hx_1+by_1+f)=x_1(ax_1+hy_1+g)+y_1(hx_1+by_1+f)\tag{5}$$
Now compare (3) and (5).
Second version, with a vectorial formulation having the advantage of more compact notations.
Expression (*) can be given the form of a quadratic equation under the form :
$$X^TQX+2V^TX+c=0 \ \text{with} \ Q=\pmatrix{a&h\\h&f}, \ \ V=\pmatrix{g\\f}, \ X=\pmatrix{x\\y}\tag{6}$$
Expression (**) becomes $X=X_1+pU$ ; plugging it into (6) gives :
$$(X_1+pU)^TQ(X_1+pU)+2V^T(X_1+pU)+c=0$$
$$X_1^TQX_1+2pU^TQX_1+p^2U^TQU+2V^TX_1+2pV^TU+c=0$$
$$p^2 \underbrace{U^TQU}_A + 2p\underbrace{U^T(QX_1+V)}_B+\underbrace{(X_1^TQX_1+c)}_C=0$$
It is exactly numerical quadratic equation (1) (check it !). In particular, setting the coefficient $2B$ of $p$ to $0$, as before, we retrieve the fact that vector $U$ is orthogonal to :
$$QX_1+V=\pmatrix{a&h\\h&f}\pmatrix{x_1\\y_1}+\pmatrix{g\\f}=\pmatrix{ax_1+hy_1+g\\hx_1+by_1+f}\tag{7}$$
as obtained in (2).
Besides, let us express $S_1$ and $T$ under their resp. vectorial expressions :
$$\begin{cases}S_1 &:& X_1^TQX_1+2 X_1^TV+c\\T &:& X^TQX_1+ (X+X_1)^TV+c\end{cases}$$
Equating them, and simplifying, we get :
$$(X-X_1)^TQX_1+(X-X_1)^TV=0$$
$$(X-X_1)^T(QX_1+V)=0$$
which is the equation of a line passing through $X=X_1$ with normal vector $QX_1+V$ as "imposed" by (7) ! QED.
Edit : I answer your question about the fact that the normal vector to the chord is "rather surprisingly"
$$QX_1+V$$
i.e., equal (up to a $2$ factor) to the gradient of expression (6) computed in point $X_1$.
Have a look at figure 2 featuring a conic curve, and a point $X_1$ which is the midpoint of a chord found by the $T=S_1$ method
Fig. 2. Homothetic curves $Q(x,y)=1$ and $Q(x,y)=1/3$ with $Q(x,y)=x^2+xy+y^2$ and $(x_1,y_1)=(1/3,1/3)$.
This chord is in fact tangent to a homothetical ellipse to the initial ellipse : no surprize that the gradient at point $X_1$ is orthogonal to the tangent to the curve.
Best Answer
This can be easily done using homogeneous representations. In a homogeneous representation, the given conic is represented as the matrix $\mathbf C$ as follows: $$ \mathbf C = \begin{bmatrix}A & H & G\\H & B & F\\G & F & C\end{bmatrix} $$
The homogeneous representation of the point $ P = (x_1, y_1) $ is,
$$p = \begin{bmatrix}x_1\\y_1\\1\end{bmatrix} $$
The homogeneous representation of the tangent line to this conic at point $P$ is, $$ l = \mathbf Cp $$ which is, $$l = \begin{bmatrix}A & H & G\\H & B & F\\G & F & C\end{bmatrix}\begin{bmatrix}x_1\\y_1\\1\end{bmatrix} = \begin{bmatrix}Ax_1+Hy_1+G\\Hx_1 + By_1+F\\Gx_1+Fy_1+C\end{bmatrix} $$
The homogeneous representation of the line $ ax+by+c=0$ is $ [a, b, c]^T$. Therefore, by multiplying the above vector $l$ with $[x, y, 1]$ you can get the given results. $$given~equation = \begin{bmatrix}Ax_1+Hy_1+G\\Hx_1 + By_1+F\\Gx_1+Fy_1+C\end{bmatrix}^T\begin{bmatrix}x\\y\\1\end{bmatrix}=0$$