Tangent to xy = c² at Parameter t

Tangent to xy = c² at (ct, c/t). Slope = −1/t². Intercepts: x-intercept 2ct, y-intercept 2c/t. The tangent at t meets the asymptotes (axes) at (2ct, 0) and (0, 2c/t), and the point (ct, c/t) is the midpoint of this segment.

At $(ct, c/t)$, slope of tangent $= -1/t^2$. Tangent equation:

$$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$$ $$t^2y - ct = -(x - ct) = -x + ct$$ $$x + t^2y = 2ct$$

Intercepts:

• $x$-intercept (set $y = 0$): $x = 2ct$, point $(2ct, 0)$
• $y$-intercept (set $x = 0$): $t^2y = 2ct \Rightarrow y = 2c/t$, point $(0, 2c/t)$

The point $(ct, c/t)$ is the midpoint of the intercept segment:

Midpoint $= \left(\frac{2ct+0}{2}, \frac{0+2c/t}{2}\right) = (ct, c/t)$ $\checkmark$

This is the tangent-midpoint property of the rectangular hyperbola — the point of tangency bisects the segment between the intercepts on the asymptotes.

Intersection of tangents at $t_1$ and $t_2$:

From $x + t_1^2y = 2ct_1$ and $x + t_2^2y = 2ct_2$:

$$(t_1^2-t_2^2)y = 2c(t_1-t_2) \implies y = \frac{2c}{t_1+t_2}$$ $$x = 2ct_1 - t_1^2 \cdot \frac{2c}{t_1+t_2} = \frac{2ct_1t_2}{t_1+t_2}$$

Intersection point: $\left(\dfrac{2ct_1t_2}{t_1+t_2},\; \dfrac{2c}{t_1+t_2}\right)$.

Note this satisfies $xy = c^2$ iff $t_1t_2 = 1$ — the intersection is on the curve only when $t_1t_2 = 1$.