Finding the Generalized Ellipse to a Tangent and the Intersecting Point under Contraints

Question

Given a line

$g: 0 = I - sX_1 - X_2$

$I, s \in R$

I need to find $V$ for an arbitrary $I$ so that there exists a point $(X_1, X_2)$ where $g_I$ is tangent to the ellipse

$V: V = X_1^2A+X_1X_2B+X_2^2C+X_1D-X_2E+F$

$A, B, C, D, E, F \in R$

More intuitively I need to find one of the concentric ellipses shown here with a solid line, so that an arbitrary line shown here dashed touches the line g and where it does so.

I have tried:

with $ V' = V+I-F$, $D' = D+s$ and $E' = E+1$ it can be shown that from g and V follows

$\implies V' = X_1^2A+X_1X_2B+X_2^2C+X_1D'-X_2E'$

Which is also an ellipse. Wikipedia provides the determinant of that ellipse which is zero when there is exactly one solution$(X_1, X_2) \in R^2$. From that follows

$V = F - \frac{B(D+s)(E+1)+a(E+1)^2+C(D+s)^2}{B^2-4AC} - I$

For the example shown this is off by about two orders of magnitude, specifically because s is two orders of magnitude larger than anything else in the central term.

Answer 1

Given an ellipse and a line as

$$ \cases{ \mathscr{E}_f\to a x^2+b x y + c y^2+ d x + e y + f=0\\ \mathscr{L}\to \iota - s x- y = 0 } $$

determine $f$ such that $\mathscr{E}_f$ and $\mathscr{L}$ are tangent. At the intersection we have that $\mathscr{E}_f$ and $\mathscr{L}$ have common $y$ then after substituting $y$ in both equations we have

$$ p(x) = a x^2+2 b \iota x-2 b s x^2+c \iota ^2-2 c \iota s x+c s^2 x^2+dx+e \iota -e s x+f = k(x-x_1)^2,\ \ \ \forall x $$

which means that at tangency, $p(x)$ has a double root. Grouping coefficients this is equivalent to

$$ \left\{ \begin{array}{rcl} c \iota ^2-k x_1^2+e \iota +f & = & 0\\ b \iota -2 c \iota s+2 k x_1+d-e s & = & 0 \\ a-b s+c s^2-k & = & 0 \\ \end{array} \right. $$

and solving for $\{k, x_1, f\}$ we obtain

$$ \left\{ \begin{array}{rcl} k&=&a-b s+c s^2 \\ x_1&=&\frac{b \iota -2 c \iota s+d-e s}{2 \left(b s-a-c s^2\right)} \\ f&=&\frac{b^2 \iota ^2+2 b d \iota +2 b e \iota s-4 c d \iota s+d^2-2 d e s+e^2 s^2-4 a c \iota ^2-4 a e \iota}{4 \left(a-b s+c s^2\right)} \\ \end{array} \right. $$

Example: Given $\{a=2,b=1,c=3,d=1,e=-1,\iota=1,s=-1\}$ we obtain $\{f = \frac {1}{24}, x_1 = -\frac{7}{12}\}$