The curve $C$ is defined parametrically by $x = \mathrm{e}^{4t}$, $y = t^2$, where $t \ge 0$

Question

The curve $C$ is defined parametrically by $x = \mathrm{e}^{4t}$, $y = t^2$, where $t \ge 0$. Find the Cartesian equation of $C$.[2] The tangent at point $P$ has the steepest gradient. Find the exact coordinates of $P$. [You do not need to show that the gradient at $P$ is the steepest.][3] Sketch $C$, indicating the coordinates of $P$ and the point where $C$ crosses the axes clearly.[2]

Accepted Answer

(i) $y=\frac{(\ln x)^2}{16}$, $x\ge1$. (ii) $P=\left(\mathrm e,\frac{1}{16}\right)$. (iii) Curve starts at the included point $(1,0)$, rises through $P$ and has no $y$-axis intercept.

2025 NJC P1 Q7

2025 NJC P1 Q7

Solution:

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