Find the line passing through the point of tangency of $ax^2+by^2+2hxy+2gx+2fy+c=0$ and an external point $P$.

Question

Question : Given a curve $$\mathcal{S} :\ ax^2+by^2+2hxy+2gx+2fy+c=0$$ and point $P(x_1,y_1)$ which is not on the curve, basically an 'external' point, find a line passing through $P$ such that it is tangent to the curve at a point . Also find the point(s) of tangency.

My Attempt :

Let the line be $$\mathcal{L}:\ \frac{y-y_1}{x-x_1}=m$$ Let the point of tangency by $(x_0,y_0)$, then differentiating the curve $$d\mathcal{S} :\ ax\ dx+by\ dy+hy\ dx+hx\ dy+ g\ dx + f\ dy =0 $$ $$\implies \frac{dy}{dx}=-\frac{ax+hy+g}{by+hx+f}$$ So slope at point of tangency or $m$ will be $$m =-\frac{ax_0+hy_0+g}{by_0+hx_0+f}$$ So we get the line as $$\mathcal{L}:\ \frac{y-y_1}{x-x_1}=-\frac{ax_0+hy_0+g}{by_0+hx_0+f}$$ $$ax_0(x-x_1)+by_0(y-y_1)+g(x-x_1)+f(y-y_1)+h(xy_0+x_0y-x_0y_1-x_1y_0)=0$$ and as $\mathcal{S}(x_0,y_0)=0$, we get $$ax_0^2+by_0^2+2hx_0y_0+2gx_0+2fy_0+c=0$$ Now what to do? How to get the points $(x_0,y_0)$?

Answer 1

Hint.

Calling $p=(x,y)'$ we can represent the conic as

$$ p'\left(\matrix{a& h\\ h& b}\right)p+2\left(g\ \ f\right)p+c=0 $$

or

$$ p'Mp+2Np+c=0 $$

now considering the line passing by $p_0 = (x_0,y_0)'$ and direction $\vec v =(v_x,v_y)'$ as

$$ p = p_0+\lambda \vec v $$

after substitution into the conic we have

$$ p_0'Mp_0+2p_0'M\vec v\lambda+\vec v'M\vec v\lambda^2+2Np_0+2N\vec v\lambda + c=0 $$

now solving for $\lambda$

$$ \lambda = \frac{(p_0'M+Np_0)\vec v\pm\sqrt{((p_0'M+Np_0)\vec v)^2-v'M\vec v(p_0'Mp_0+2Np_0+c)}}{\vec v'M\vec v} $$

As the line should be tangent we need

$$ ((p_0'M+Np_0)\vec v)^2-v'M\vec v(p_0'Mp_0+2Np_0+c)=0 $$

Now solving

$$ \cases{ ((p_0'M+Np_0)\vec v)^2-v'M\vec v(p_0'Mp_0+2Np_0+c)=0\\ \|\vec v\|=1 } $$

two equations and two unknowns, we are done.

Answer 2

By Joachimsthal $a x x_1+b y y_1+h (x_1 y+y_1 x)+g (x+x_1)+f (y+y_1)+c=0$ is the polar (read: the line passing through the points of tangency of the conic and an external point $P=(x_1,y_1)$). To get the tangent points you need to solve the system $$ ax^2+by^2+2hxy+2gx+2fy+c=0\\a x x_1+b y y_1+h (x_1 y+y_1 x)+g (x+x_1)+f (y+y_1)+c=0$$ Which has solutions you get from the quadratic formula on $I$ and substitute into $II$ in the following (found by taking the lexicographic grobner basis): $$I: Ay^2+By+C=0,\\ II: (ax_1+hy_1+g)x+(hx_1+by_1+f)y+gx_1+fy_1+c=0,\\A=((a^2b-ah^2)x_1^2+(2abh-2h^3)x_1y_1+(ab^2-bh^2)y_1^2+(2abg-2h^2g)x_1+(2abf-2h^2f)y_1+af^2-2hfg+bg^2),\\B=(4afg-4hg^2)x_1+(2af^2-2bg^2+2abc-2h^2c)y_1+2afc-2hgc),\\C=((2a^2f-2ahg)x_1^2+(4ahf-4h^2g)x_1y_1+(2abf-2bhg)y_1^2+By+(-ag^2+a^2c)x_1^2+(-2hg^2+2ahc)x_1y_1+(af^2-2hfg+h^2c)y_1^2+(-2g^3+2agc)x_1+(-2fg^2+2afc)y_1-g^2c+ac^2.$$

Answer 3

In general, a line with slope $m$ passing through a point $(x_1,y_1)$ will have an equation of the form $y-y_1=m(x-x_1)$. We found that for $\mathcal{S}$, $\frac{dy}{dx}=-\frac{ax+hy+g}{by+hx+f}$ is the slope of the line tangent to $\mathcal{S}$ at the point $(x,y)$, giving overall the equation for a line passing through the point $P(x_1,y_1)$, tangent to $\mathcal{S}$ at the point $(x_0,y_0)$: \begin{equation} y-y_1=-\frac{ax_0+hy_0+g}{by_0+hx_0+f}(x-x_1)\text{.} \end{equation} The point $(x_0,y_0)$ lies on both $\mathcal{S}$ and the tangent line, therefore it must satisfy their respective equations, giving overall the system of equations \begin{cases} y_0-y_1=-\frac{ax_0+hy_0+g}{by_0+hx_0+f}(x_0-x_1)\\ ax_0^2+by_0^2+2hx_0y_0+2gx_0+2fy_0+c=0\text{,} \end{cases} which we solve for $(x_0,y_0)$. Can you take it from here?

Answer 4

I also found one way to find $(x_0,y_0)$, now as it turns out doing this with a derivative is redundant, alternatively we can simply put $$y = y_1+m(x-x_1)$$ in the curve from which we get $$ax^2 + b(y_1+m(x-x_1))^2+2h(y_1+m(x-x_1))x+2gx+2f(y_1+m(x-x_1))+c=0$$ $$x^2(a+m^2b+2hm)+x(2fm+2g+2bmy_1-2m^2bx_1+2hy_1-2hx_1)+by_1^2+b^2m^2x_1^2-2bmx_1y_1+2fy_1-2fmx_1+c=0$$ Now for tangency its discriminant must be zero so we get $$(fm+g+bmy_1-m^2bx_1+hy_1-hx_1)^2 - (a+m^2b+2hm)(by_1^2+b^2m^2x_1^2-2bmx_1y_1+2fy_1-2fmx_1+c)=0 $$ now we can solve for $m$ through here put in back in the OG expression and find $(x_0,y_0).$

Answer 5

The given equation of the conic can be written as

$ F(r) = r^T Q r + \ell^T r + c = 0 \tag{1} $

where $r = [x, y]^T $, $Q = \begin{bmatrix} a && h \\ h && b \end{bmatrix} $, $ \ell = \begin{bmatrix} 2g \\ 2f \end{bmatrix}$

Now, the gradient vector of the quadratic function $F(r)$ in $(1)$ is given by

$\nabla_r F = 2 Q r + \ell \tag{2}$

This gradient vector points in a direction that is perpendicular to the curve of the conic specified by $(1)$. Hence, we want to find $r$ such that the vector $\nabla_r F$ is perpendicular to the vector $(r - P)$, that is we want to solve

$ (r - P)^T (2 Q r + \ell) = 0 \tag{3}$

Taking the system of equations $(1)$ and $(3)$ we can solve for $r$ satisfying both. Here's how to do just that,

Expanding equation $(3)$, we get

$ 2 r^T Q r + r^T \ell - 2 r^T Q P - P^T \ell = 0 \tag{4}$

Using $(1)$, we have $ r^T Q r = - \ell^T r - c $, therefore, substituting this into $(4)$, gives us,

$ - r^T ( \ell + 2 Q P ) - P^T \ell = 0 \tag{5}$

This is an equation of a straight line, because it is of the form

$ A x + B y = C \tag{6}$

Define $V = [- B, A ]^T $ and $r_0 = [C/A, 0]^T $ if $A \ne 0$ or $r_0 = [0, C/B]^T$ if $A = 0$, then this line is parametrically given by

$ r = [x, y]^T = r_0 + t \ V \tag{7}$

Substitute this into $(1)$, then

$(r_0 + t \ V)^T Q (r_0 + t \ V) + \ell^T (r_0 + t \ V) + c = 0 \tag{8}$

Expanding,

$ t^2 (V^T Q V) + t ( 2 V^T Q r_0 + \ell^T V ) + (r_0^T Q r_0 + \ell^T r_0 + c ) = 0 \tag{9}$

Solving using the quadratic formula, gives two values for $t$. Substituting these into $(7)$ gives the required points of tangency (There can be a maximum of two such points).