Im having problem in many questions of this type - "Find the common tangent to the given curves". Usually the curves given are shifted standard conics.
I have two approaches for this-
Assume a line y = mx + c, substitute y in the conics, since they intersect at one point only; make the Discriminant zero and find c and m.
My other approach is using standard equations for tangents which we were taught in class.
The problem is these standard eq. Only apply for standard conics and im having problem in the rotated versions of the conics.
Please help me.
(Im new here. Please tell me if im breaking any rules)
One general approach to finding common tangents of a pair of conics is to solve the dual problem: compute the intersections of the dual conics.
If you represent tangent lines to a conic by equations of the form $\lambda x+\mu y+\tau = 0$, it turns out that the coefficients of these equations themselves satisfy a general conic equation. This is the dual conic. If you represent the original conic as a matrix equation $$\mathbf x^T C \mathbf x = 0$$ then the matrix of the dual conic is the adjugate of $C$. For non-degenerate conics, for which $C$ is nonsingular, you can also use $C^{-1}$.
Solving this intersection problem involves solving a cubic equation in the general case, so I encourage you to take advantage of the geometry of specific problems that you’re trying to solve in order to simplify the calculations.
The general conic equation $ax^2+2bxy+cy^2+2dx+2ey+f=0$ has the corresponding matrix $$C = \begin{bmatrix}a&b&d\\b&c&e\\d&e&f\end{bmatrix}.$$ Its adjugate is $$C^{\tiny\triangle} = \begin{bmatrix} cf-e^2 & de-bf & be-cd \\ de-bf & af-d^2 & bd-ae \\ be-cd & bd-ae & ac-b^2\end{bmatrix}.$$ Basically, this is the matrix of $2\times2$ submatrix determinants multiplied by $(-1)^{i+j}$, where $i$ and $j$ are the respective row and column indices. So, if the equation of a tangent line to this conic is $\lambda x+\mu y+\tau = 0$, its coefficients satisfy the equation $$\begin{bmatrix}\lambda&\mu&\tau\end{bmatrix}C^{\tiny\triangle}\begin{bmatrix}\lambda\\\mu\\\tau\end{bmatrix} = (cf-e^2)\lambda^2 + (af-d^2)\mu^2 + (ac-b^2)\tau^2 + 2(de-bf)\lambda\mu + 2(be-cd)\lambda\tau + 2(bd-ae)\mu\tau = 0.$$