Deduce the y-coordinate of circumcentre of a triangle involving completing square and quadratic function transformation

Question Sample Titled 'Deduce the y-coordinate of circumcentre of a triangle involving completing square and quadratic function transformation'

Let ${f{{\left({x}\right)}}}={2}{x}^{{2}}-{4}{k}{x}+{6}{k}^{{2}}+{4}$ , where ${k}$ is a real constant.

 (a) Does the graph of ${y}={f{{\left({x}\right)}}}$ cut the ${x}$-axis? Explain your answer. (2 marks) (b) Using the method of completing the square, express, in terms of ${k}$ , the coordinates of the vertex of the graph of ${y}={f{{\left({x}\right)}}}$ . (3 marks) (c) In the same rectangular coordinate system, let ${S}$ and ${T}$ be moving points on the graph of ${y}={f{{\left({x}\right)}}}$ and the graph of ${y}={3}-{f{{\left({x}\right)}}}$ respectively. Denote the origin by ${O}$ . Someone claims that when ${S}$ and ${T}$ are nearest to each other, the circumcentre of $\triangle{O}{S}{T}$ lies on the ${x}$-axis. Is the claim correct? Explain your answer. (4 marks)

 (a) Discriminant ${\left(\Delta\right)}$ of ${f{{\left({x}\right)}}}$ $={\left(-{4}{k}\right)}^{{2}}-{4}{\left({2}\right)}{\left({6}{k}^{{2}}+{4}\right)}$ 1M $={16}{k}^{{2}}-{48}{k}^{{2}}-{32}$ $=-{32}{k}^{{2}}-{32}$ accepts $-{32}{\left({k}^{{2}}+{1}\right)}$ $\lt{0}$ Thus, the graph of ${y}={f{{\left({x}\right)}}}$ does not cut the ${x}$-axis. 1A f.t. (b) ${f{{\left({x}\right)}}}$ $={2}{x}^{{2}}-{4}{k}{x}+{6}{k}^{{2}}+{4}$ $={2}{\left({x}^{{2}}-{2}{k}{x}\right)}+{6}{k}^{{2}}+{4}$ $={2}{\left({x}^{{2}}-{2}{k}{x}+{k}^{{2}}-{k}^{{2}}\right)}+{6}{k}^{{2}}+{4}$ 1M for completing the perfect square $={2}{\left({x}-{k}\right)}^{{2}}+{4}{k}^{{2}}+{4}$ 1A Thus, the coordinates of the vertex are ${\left({k},{4}{k}^{{2}}+{4}\right)}$ . 1M (c) By (b), the coordinates of the vertex of the graph of ${y}=-{f{{\left({x}\right)}}}$ are ${\left({k},-{4}{k}^{{2}}-{4}\right)}$ . So, the coordinates of the vertex of the graph of ${y}={3}-{f{{\left({x}\right)}}}$ are ${\left({k},-{4}{k}^{{2}}-{1}\right)}$ . 1M When ${S}$ and ${T}$ are nearest to each other, the coordinates of ${S}$ and ${T}$ are ${\left({k},{4}{k}^{{2}}+{4}\right)}$ and ${\left({k},-{4}{k}^{{2}}-{1}\right)}$ respectively. Notice that ${S}{T}$ is a vertical line. 1M So, the perpendicular bisector of ${S}{T}$ is a horizontal line. The ${y}$-coordinate of the circumcentre of $\triangle{O}{S}{T}$ $=\dfrac{{{\left({4}{k}^{{2}}+{4}\right)}+{\left(-{4}{k}^{{2}}-{1}\right)}}}{{2}}$ 1M $=\dfrac{{3}}{{2}}$ $\ne{0}$ Therefore, the circumcentre of $\triangle{O}{S}{T}$ does not lie on the ${x}$-axis. Thus, the claim is incorrect. 1A f.t. Refer to the figure below, which shows one possible situation.

${x}$${y}$${S}{\left({k},{4}{k}^{{2}}+{4}\right)}$${T}{\left({k},-{4}{k}^{{2}}-{1}\right)}$${y}={f{{\left({x}\right)}}}$${y}={3}-{f{{\left({x}\right)}}}$

 (c) Assume that when ${S}$ and ${T}$ are nearest to each other, the circumcentre of $\triangle{O}{S}{T}$ lies on the ${x}$-axis. The coordinates of ${S}$ and ${T}$ are ${\left({k},{4}{k}^{{2}}+{4}\right)}$ and ${\left({k},-{4}{k}^{{2}}-{1}\right)}$ respectively. 1M Let ${\left({c},{0}\right)}$ be the coordinates of the circumcentre ${R}$ of $\triangle{O}{S}{T}$ . ${R}{S}$ 1M either one $=\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({0}-{\left({4}{k}^{{2}}+{4}\right)}\right)}^{{2}}}}$ $=\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({4}{k}^{{2}}+{4}\right)}^{{2}}}}$ ${R}{T}$ $=\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({0}-{\left(-{4}{k}^{{2}}-{1}\right)}\right)}^{{2}}}}$ $=\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({4}{k}^{{2}}+{1}\right)}^{{2}}}}$ So, solving for ${k}$ , ${R}{S}$ $={R}{T}$ $\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({4}{k}^{{2}}+{4}\right)}^{{2}}}}$ $=\sqrt{{{\left({c}-{k}\right)}^{{2}}+{\left({4}{k}^{{2}}+{1}\right)}^{{2}}}}$ ${\left({4}{k}^{{2}}+{4}\right)}^{{2}}$ $={\left({4}{k}^{{2}}+{1}\right)}^{{2}}$ ${\left({4}{k}^{{2}}+{4}\right)}^{{2}}-{\left({4}{k}^{{2}}+{1}\right)}^{{2}}$ $={0}$ ${\left({8}{k}^{{2}}+{5}\right)}{\left({3}\right)}$ $={0}$ ${k}^{{2}}$ $=-\dfrac{{5}}{{8}}$ Contradiction Since ${k}$ is a real constants, ${R}{S}\ne{R}{T}$ . 1M Therefore, the circumcentre of $\triangle{O}{S}{T}$ is not possible to be on the ${x}$-axis. Thus, the claim is incorrect. 1A f.t.

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