Help finding the tangent line to a curve

Question

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?

Answer 1

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. $$given~equation = \begin{bmatrix}Ax_1+Hy_1+G\\Hx_1 + By_1+F\\Gx_1+Fy_1+C\end{bmatrix}^T\begin{bmatrix}x\\y\\1\end{bmatrix}=0$$

Answer 2

Given a level set $M=\{(x,y)\in\mathbb R^2 : f(x,y)=0\}$, the gradient $\nabla f=\binom{\partial_x f}{\partial_y f} $ of $f$, if it does not vanish, will point in the direction of steepest ascent, and therefore be normal to $M$. The tangent line at $(x_1,y_1)$ is precisely the set of points $(x,y)$ such that the vectors $(x-x_1,y-y_1)$ are perpendicular to a normal vector at $(x_1,y_1)$, i.e. they are solutions to the equation $$ \nabla f (x_1,y_1)\cdot \binom{x-x_1}{y-y_1} = 0. \tag{*}\label{*}$$ In your case, $$\nabla f(x_1,y_1) = \binom{2Ax_1+2Hy_1+2G}{2By_1+2Hx_1+2F},$$ so the LHS of \eqref{*} is \begin{align} \nabla f(x_1,y_1) \cdot \binom{x-x_1}{y-y_1} &= \nabla f (x_1,y_1)\cdot \binom{x}{y} - \nabla f (x_1,y_1)\cdot \binom{x_1}{y_1} \\&=2Ax_1x+2Hy_1x+2Gx+2By_1y+2Hx_1y+2Fy \\ &\quad - (2Ax_1^2+2Hy_1x_1+2Gx_1+2By_1^2+2Hx_1y_1+2Fy_1) \\&= 2Ax_1x+2Hy_1x+2Gx+2By_1y+2Hx_1y+2Fy \\ &\quad - \color{blue}{(2Ax_1^2+4Hy_1x_1+4Gx_1+2By_1^2+4Fy_1+2C)} \\&\quad \phantom{- (2Ax_1^2+4Hy_1x_1+} + 2G x_1 \phantom{{}+2By_1^2} + 2Fy_1 + 2C \\&= 2Ax_1x+2Hy_1x+2Gx+2By_1y+2Hx_1y+2Fy \\ &\quad - \color{blue}{2f(x_1,y_1)} \\&\quad + 2G x_1 + 2Fy_1 + 2C \\ &=2Ax_1x+2H(y_1x+x_1y)+2G(x+x_1)+2By_1y+2F(y+y_1) + 2C. \end{align} We added and subtracted terms to use that $x_1,y_1$ satisfy $f(x_1,y_1)=0$. Cancelling the factor of 2 and rearranging, we obtain the desired equation. It holds so long as $\binom{2Ax_1+2Hy_1+2G}{2By_1+2Hx_1+2F}$ is not the zero vector.