Find the point where pair of tangents intersect What is the point of intersection of tangents drawn to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the points where it is intersected by the line $lx+my+n=0$ ?
$(A)$ $\big(-\frac{a^2l}{n},\frac{b^2m}{n} \big)$
$(B)$ $\big(-\frac{a^2l}{m},\frac{b^2n}{m} \big)$
$(C)$ $\big(\frac{a^2l}{m},-\frac{b^2n}{m} \big)$
$(D)$ $\big(\frac{a^2l}{m},\frac{b^2n}{m} \big)$
Solving line and hyperbola and then finding required point is turning out to be too tedious. Could someone suggest a better approach?
 A: It is solution (A).
This question is typical of projective geometry, more exactly of its subbranch called the theory of duality (or conjugation) between points  (called "poles") and straight lines (called "polars") with respect to a conical section. (see the following book "Symmetry and Pattern in Projective Geometry", Eric Lord, Springer, 2012). I know that it is not very easy to enter into the "realm" of projective geometry but knowing its existence and how can a problem be solved in this framework is a first objective. A deep understanding of the subject is not needed here.
Definition : $(x,y,t)$ and $(x',y',t')$ are said to be conjugated with respect to conical section $x^2/a^2-y^2/b^2-t^2=0$ (we have a supplementary coordinate $t$ whose value is 1 for "ordinary points", and $0$ for "points at infinity") iff:
$$xx'/a^2-yy'/b^2-tt'=0 \Longleftrightarrow \ \begin{pmatrix}x&y&t\end{pmatrix}\begin{pmatrix}1/a^2&&\\&-1/b^2&\\&&-1\end{pmatrix}\begin{pmatrix}x'\\y'\\t'\end{pmatrix}=0$$
Remark: Consider $xx'/a^2-yy'/b^2-tt'$  as a kind of generalization of the Euclidean dot product $xx'+yy'$ associated with quadratic form $x^2+y^2$ ; and the fact that this quantity is $0$ generalizes condition of Euclidean orthogonality $xx'+yy'=0 $).
The polar of a certain pole with projective coordinates $P_0(x_0,y_0,t_0)$ is the set of points that are conjugated to it, i.e. with equation
$$\begin{pmatrix}x&y&t\end{pmatrix}\begin{pmatrix}1/a^2&&\\&-1/b^2&\\&&-1\end{pmatrix}\begin{pmatrix}x_0\\y_0\\t_0\end{pmatrix}=0$$
Developing:
$$x(x_0/a^2)+y(-y_0/b^2)-tt_0=0 $$
Replacing $t$ and $t_0$ by $1$ because the points consider here are at finite distance: 
$$\tag{1}x(x_0/a^2)+y(-y_0/b^2)-1=0$$ 
Equation (1) must be the same as equation 
$$\tag{2}\ell x + m y + n = 0,$$
which amounts to say that their coefficients are proportionnal:
$$\dfrac{(x_0/a^2)}{\ell}=\dfrac{(-y_0/b^2)}{m}=\dfrac{-1}{n}$$
$$\Rightarrow \ \ x_0=-a^2 \ell/n, \ \ y_0=b^2 m/n$$
proving that it is solution (A).
A: A shortcut, if you can rely on the answer being one of the four, is to observe that when $n = 0$, the intersecting line cuts through the origin, and therefore the tangent lines are parallel and never intersect.  That is consistent only with answer (A).
