A curve has equation $y = \mathrm{f}(x)$, where $\mathrm{f}(x) = 1 - \sqrt{q^2 - x^2}$ for&

Question

A curve has equation $y = \mathrm{f}(x)$, where $\mathrm{f}(x) = 1 - \sqrt{q^2 - x^2}$ for $q > 1$. State the shape of $y = \mathrm{f}(x)$.[1] Sketch the curve $y = \frac{1}{\mathrm{f}(x)}$, giving the equations of any asymptotes and the coordinates of the end-points.[3] Describe the transformations that map the graph of $y = \mathrm{f}(x)$ to $y = -\sqrt{1-x^2}$.[3]

Accepted Answer

The graph of $y=f(x)$ is the lower semicircle with centre $(0,1)$ and radius $q$. (i) For $y=1/f(x)$, the vertical asymptotes are $x=\pm\sqrt{q^2-1}$ and the endpoints are $(-q,1)$ and $(q,1)$. (ii) Translate $1$ unit down, scale parallel to the $y$-axis by factor $1/q$, then scale parallel to the $x$-axis by factor $1/q$.

2025 NJC P1 Q4

2025 NJC P1 Q4

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