From any point of the hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$, tangents are drawn to another hyperbola which has the same asymptotes. Show that the chord of contact cuts off a constant area from the asymptotes.
Concept
Lets presume that hyperbola are conjugate
Let the two conjugate hyperbolas be represented as
$\ H_1$: $\frac{y^2}{b^2} -\frac{x^2}{a^2}=1$
$\ H_2$: $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$
There Asymptotes are
$\ A_1$: $\frac{y}{b} -\frac{x}{a}=1$
$\ A_2$: $\frac{x}{a} -\frac{y}{b}=1$
Points on $\ H_1$ is $(a\tan(\alpha), b\sec(\alpha))$
Points on $\ H_2$ is $(a\sec(\alpha), b\tan(\alpha))$
Equation of Line tangent to $\ H_1$
$\frac{y\sec(\alpha)}{b}-\frac{x\tan(\alpha)}{a}$=1, it passes through $(a\sec(\alpha), b\tan(\alpha))$
From here i need help. Do we need to find the intersection of the line and asymptote and then find area under curve that needs to be "Constant"
If i enter coordinates of $\ H_2$ in equation of line equation vanishes
You need to find the line through the points of tangency, its intersections with the common asymptotes and then the area of the resulting triangle. This is a fairly straightforward calculation if you use homogeneous coordinates.
The family of hyperbolas with the same asymptotes as ${x^2\over a^2}-{y^2\over b^2}=1$ is given by the equations ${x^2\over a^2}-{y^2\over b^2}=k$. (For $k=0$, this is the equation of the asymptotes themselves.) In matrix form, these hyperbolas are $H=\operatorname{diag}(1/a^2,-1/b^2,-k)$. The line of contact of the tangents to a hyperbola from a point $(x,y)$ is the polar line of this point, namely, $$\mathbf l=H[x:y:1]=\left[{x\over a^2}:-{y\over b^2}:-k\right].$$ The two asymptotes are $\mathbf m_1=[b:a:0]$ and $\mathbf m_2=[b:-a:0]$, and their intersections with $\mathbf l$ are $$ \mathbf p_1=\mathbf l\times\mathbf m_1=\left[ak:-bk:\frac xa+\frac yb \right] \\ \mathbf p_2=\mathbf l\times\mathbf m_2=\left[-ak:-bk:-\frac xa+\frac yb\right] $$ which in Cartesian coordinates are, respectively, $$ P_1=\left({a^2bk\over bx+ay},-{ab^2k\over bx+ay}\right) \\ P_2=\left({a^2bk\over bx-ay},{ab^2k\over bx-ay}\right).$$ The area of $\triangle{OP_1P_2}$ is one half of the absolute value of $${a^2bk\over bx+ay}{ab^2k\over bx-ay}+{ab^2k\over bx+ay}{a^2bk\over bx-ay} = 2{a^3b^3k^2\over b^2x^2-a^2y^2}.$$ Multiplying the equation of the original hyperbola by $a^2b^2$ gives the equivalent equation $b^2x^2-a^2y^2=a^2b^2$, so for any point on this hyperbola the area of $\triangle{OP_1P_2}$ is equal to $abk^2$.
Observe that this didn’t use the points of tangency directly but instead used the polar line of a point, so this identity also holds when the point is interior to the other hyperbola, in which case there are no (real) tangent lines, and when the two hyperbolas coincide, in which case there’s only one point of tangency. Also, for any hyperbola in the family $b^2x^2-a^2y^2$ is constant, therefore this constant-area property holds for any pair of hyperbolas with common asymptotes.