Graphs & Transformations: Reciprocal & transformation description — NJC 2025 H2 Math Prelim Paper 1
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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 \(y=\mathrm{f}(x)\).
- Sketch the curve \(y=\dfrac{1}{\mathrm{f}(x)}\), giving the equations of any asymptotes and the coordinates of the end-points.
- Describe the transformations that map the graph of \(y=\mathrm{f}(x)\) to \(y=-\sqrt{1-x^2}\).
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\(y = \mathrm{f}(x) = 1 - \sqrt{q^2-x^2}\).
Rearranging: \((y-1)^2 = q^2-x^2\), so \(x^2+(y-1)^2=q^2\).
\(y = \mathrm{f}(x)\) is the lower half of the circle centred at \((0,1)\) with radius \(q\) (since \(\sqrt{q^2-x^2}\geq 0\) implies \(y\leq 1\)).
(i)
For \(y=\dfrac{1}{\mathrm{f}(x)}\): \(\mathrm{f}(x)=0\) when \(1-\sqrt{q^2-x^2}=0\), i.e. \(x=\pm\sqrt{q^2-1}\).
Vertical asymptotes: \(x=\pm\sqrt{q^2-1}\). End-points: \((-q,1)\) and \((q,1)\).
(ii)
Starting from \(y=\mathrm{f}(x) = 1-\sqrt{q^2-x^2}\), target is \(y=-\sqrt{1-x^2}\).
Step 1: Translate 1 unit in the negative \(y\)-direction: \(y \mapsto y+1\), giving \(y = -\sqrt{q^2-x^2}\).
Step 2: Scale by factor \(\dfrac{1}{q}\) parallel to the \(x\)-axis: \(x \mapsto qx\), giving \(y = -\sqrt{q^2-q^2x^2} = -q\sqrt{1-x^2}\).
Step 3: Scale by factor \(\dfrac{1}{q}\) parallel to the \(y\)-axis: \(y \mapsto qy\), giving \(y = -\sqrt{1-x^2}\).