Tangent–Asymptote Triangle

The tangent at any point P on the hyperbola meets the two asymptotes at Q and R. P is the midpoint of QR, and the area of triangle OQR (O = centre) is constant = ab, independent of the choice of P.

Let $P(a\sec\theta, b\tan\theta)$ be a point on $x^2/a^2 - y^2/b^2 = 1$. The tangent at $P$ is:

$$\frac{x\sec\theta}{a} - \frac{y\tan\theta}{b} = 1$$

The asymptotes are $L_1: bx - ay = 0$ and $L_2: bx + ay = 0$.

Intersection with $L_1$ (i.e., $y = bx/a$): Substitute into tangent:

$$\frac{x\sec\theta}{a} - \frac{(bx/a)\tan\theta}{b} = 1 \implies \frac{x}{a}(\sec\theta - \tan\theta) = 1 \implies x = \frac{a}{\sec\theta-\tan\theta}$$ $$Q = \left(\frac{a}{\sec\theta-\tan\theta},\; \frac{b}{\sec\theta-\tan\theta}\right)$$

Intersection with $L_2$ (i.e., $y = -bx/a$):

$$R = \left(\frac{a}{\sec\theta+\tan\theta},\; \frac{-b}{\sec\theta+\tan\theta}\right)$$

P is the midpoint of QR: Midpoint x-coordinate:

$$\frac{1}{2}\!\left(\frac{a}{\sec\theta-\tan\theta}+\frac{a}{\sec\theta+\tan\theta}\right) = \frac{a}{2}\cdot\frac{2\sec\theta}{\sec^2\theta-\tan^2\theta} = a\sec\theta \checkmark$$

Similarly the y-coordinate gives $b\tan\theta$. So $P$ is the midpoint of $QR$.

Area of triangle $OQR$:

$$\text{Area} = \frac{1}{2}|OQ \times OR| = \frac{1}{2}\left|\frac{a}{\sec\theta-\tan\theta}\cdot\frac{-b}{\sec\theta+\tan\theta} - \frac{b}{\sec\theta-\tan\theta}\cdot\frac{a}{\sec\theta+\tan\theta}\right|$$ $$= \frac{1}{2}\cdot\frac{2ab}{\sec^2\theta-\tan^2\theta} = \frac{ab}{1} = ab$$

The area is $ab$ — constant, independent of $P$.