2015 HCI Promo Q2

2015 HCI Promo Q2

6 marks
Promo

The curve \(C\) has equation \(4{{x}^{2}}+{{y}^{2}}+8mx-4y+4=0\), where \(m\in \mathbb{R}\) , \(m>0\).

Show that \(C\)can be expressed in the form \({{\left( \frac{x+a}{b} \right)}^{2}}+{{\left( \frac{y+c}{d} \right)}^{2}}=1\), where \(a\) , \(b\) , \(c\) and \(d\) are constants to be determined in terms of \(m\) .

Hence describe a sequence of transformations, in terms of \(m\), which transforms the graph of \({{x}^{2}}+{{\left( y-\frac{1}{m} \right)}^{2}}=1\) to \(C\).[6]

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Answer:\(C: \left(\dfrac{x+m}{m}\right)^2+\left(\dfrac{y-2}{2m}\right)^2=1\), with \(a=m\), \(b=m\), \(c=-2\), \(d=2m\) One sequence: scale parallel to the \(x\)-axis by \(m\), translate \(m\) units left, then scale parallel to the \(y\)-axis by \(2m\)

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