### quadratic transform, circumcentre through vertices

Question Sample Titled 'quadratic transform, circumcentre through vertices'

Let ${f{{\left({x}\right)}}}={3}{x}^{{2}}-{6}{a}{x}+{3}{a}^{{2}}+{4}$ , where ${a}$ is a non-zero 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 ${a}$, the coordinates of vertex of the graph ${y}={f{{\left({x}\right)}}}$ . (3 marks) (c) On the same rectangular coordinate plane, let ${H}$ and ${K}$ be two moving points on the graph of ${y}={f{{\left({x}\right)}}}$ and the graph of ${y}=-{f{{\left({x}\right)}}}-{3}$ respectively. Denote the origin by ${O}$. Holly claims that when ${H}$ and ${K}$ are closest to each other, the circumcentre of $\triangle{H}{O}{K}$ lies on the ${x}$-axis. Is she correct? Explain your answer. (4 marks)

 (a) Consider ${f{{\left({x}\right)}}}$ $={0}:$ $\Delta$ $={\left(-{6}{a}\right)}^{{2}}-{4}{\left({3}\right)}{\left({3}{a}^{{2}}+{4}\right)}$ 1M  $={36}{a}^{{2}}-{36}{a}^{{2}}-{48}$  $=-{48}$  $<{0}$ 1A ∴  The graph does not cut the ${x}$-axis.  (b) ${f{{\left({x}\right)}}}$ $={3}{x}^{{2}}-{6}{a}{x}+{3}{a}^{{2}}+{4}$  $={3}{\left({x}^{{2}}-{2}{a}{x}\right)}+{3}{a}^{{2}}+{4}$  $={3}{\left({x}^{{2}}-{2}{a}{x}+{a}^{{2}}-{a}^{{2}}\right)}+{3}{a}^{{2}}+{4}$ 1M  $={3}{\left({x}-{a}\right)}^{{2}}-{3}{a}^{{2}}+{3}{a}^{{2}}+{4}$  $={3}{\left({x}-{a}\right)}^{{2}}+{4}$ 1A ∴  Vertex $={\left({a},{4}\right)}$ 1A  (c) The vertex of the graph of ${y}=-{f{{\left({x}\right)}}}-{3}$ is ${\left({a},-{7}\right)}$ 1A ${H}$ and ${K}$ are closest to each other when they are the vertex of ${y}={f{{\left({x}\right)}}}$ and ${y}=-{f{{\left({x}\right)}}}-{3}$ respectively. 1M Since the ${x}$-coordinates of these two vertices are the same and because radius of circle is the perpendicular bisector of chord, ∴  The ${y}$-coordinate of the circumcentre of $\triangle{H}{O}{K}$ is equal to the ${y}$-coordinate of the mid-point of ${H}$ and ${K}$ . ${y}$-coordinate of the mid-point of ${H}$ and ${K}=\dfrac{{{4}+{\left(-{7}\right)}}}{{2}}$ $={1.5}$ 1A $\ne{0}$ ∴  The circumcentre of $\triangle{H}{O}{K}$ does not lie on the ${x}$-axis. Thus, she is not correct. 1A

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