<s>[INST] Pair of tangents to a hyperbola How do I find the joint equation of the pair of tangents drawn to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ from an external point $(x_1, y_1)$.
My book says that the answer is $SS_1 = T^2$, where
$$
S = \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1\\
S_1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1\\
T = \frac{x x_1}{a^2} - \frac{y y_1}{b^2} - 1
$$
But I can't get this result. Please help me go about this problem.
 [/INST] Let $P(x_1,y_1)$, and let $P'(\alpha,\beta)$ be any point in the plane. Let $V(x',y')$ lie on the line $PP'$ so that the ratio $PV:VP'=m:n$.
Then $$V(x',y')=\left(\frac{m\alpha+nx_1}{m+n},\frac{m\beta+ny_1}{m+n}\right)$$
Now let $V$ lie on the hyperbola, so that $$\left(\frac{m\alpha+nx_1}{a(m+n)}\right)^2-\left(\frac{m\beta+ny_1}{b(m+n)}\right)^2=1$$
This can be rearranged, after a couple of lines, to form a quadratic in $\frac mn$, namely, $$\left(\frac mn\right)^2\left(b^2\alpha^2-a^2\beta^2-a^2b^2\right)+2\left(\frac mn\right)\left(b^2\alpha x_1-a^2\beta y_1-a^2b^2\right)+\left(b^2x_1^2-a^2y_1^2-a^2b^2\right)=0$$
Now if $P'$ lies on either of the tangents to the hyperbola, this quadratic has double roots, which gives us $$\left(b^2\alpha^2-a^2\beta^2-a^2b^2\right)^2=\left(b^2\alpha x_1-a^2\beta y_1-a^2b^2\right)\left(b^2x_1^2-a^2y_1^2-a^2b^2\right)$$
Now replace $(\alpha, \beta)$ with $(x,y)$ and divide throughout by $\left(a^2b^2\right)^2$ and you get the stated result $SS_1=T^2$
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