A curve $C$ is defined parametrically by the equations \[x = t + \frac{1}{t}, \quad y = t - \frac{

Question

A curve $C$ is defined parametrically by the equations $$x = t + \frac{1}{t}, \quad y = t - \frac{1}{t}, \quad t \neq 0.$$ Sketch the curve $C$, showing clearly the coordinates of the axial intercepts.[2] Use differentiation to find the values of $t$ for which the tangents to the curve are parallel to the $y$-axis.[3] Show that the equation of normal at the point where $t = 2$ is given by $y = -\frac{3}{5}x + 3$.[3] The normal at the point where $t = 2$ cuts the curve $C$ again at the point $Q$. Determine the exact coordinates of $Q$.[3]

Accepted Answer

(a) $x^2-y^2=4$, axial intercepts $(2,0)$, $(-2,0)$; no $y$-intercept. (b) $t=\pm1$. (c) $y=-\frac35x+3$ (shown). (d) $Q=\left(-\frac{65}{8},\frac{63}{8}\right)$.

2025 TJC P1 Q9

2025 TJC P1 Q9

Solution:

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