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Conic Sections and Parametric Equations: math test solutions, Grade 12 – download the PDF

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Test solutions Grade 12 : Conic Sections and Parametric Equations — Zyro the alien explorer of Planète Maths

Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

1 Circle / 4 pts

  1. \((x^2+6x)+(y^2-10y)=-18\), so \((x+3)^2+(y-5)^2=-18+9+25=16\). Center \((-3,5)\), radius \(4\). (2 pts)
  2. \((0)^2+(9-5)^2=16\): yes, the point is on the circle. (1 pt)
  3. With \(y=5\): \((x+3)^2=16\), so \(x=1\) or \(x=-7\). Points \((1,5)\) and \((-7,5)\). (1 pt)

2 Parabola / 4 pts

  1. \(4p=16\), \(p=4\). Vertex \((0,0)\), focus \((4,0)\), directrix \(x=-4\). (2 pts)
  2. \(y^2=144\), so \(y=\pm12\): points \((9,\pm12)\) (1 pt). Distance to the focus: \(\sqrt{(9-4)^2+12^2}=\sqrt{169}=13\); distance to the directrix: \(9+4=13\). (1 pt)

3 Ellipse / 4 pts

  1. Divide by \(400\): \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\). \(a=5\), \(b=4\), \(c=\sqrt{25-16}=3\), vertices \((\pm5,0)\). (2 pts)
  2. Foci \((\pm3,0)\); \(e=\dfrac{3}{5}\). (1 pt)
  3. \(PF_2=\dfrac{16}{5}=3.2\) (same \(x\) as the focus \((3,0)\)); \(PF_1=\sqrt{6^2+3.2^2}=\sqrt{46.24}=6.8\). Sum \(=10=2a\). (1 pt)

4 Hyperbola / 4 pts

  1. \(a=6\), \(b=8\), \(c=\sqrt{36+64}=10\), \(e=\dfrac{10}{6}=\dfrac{5}{3}\). (2 pts)
  2. The branches open up and down: vertices \((0,\pm6)\), foci \((0,\pm10)\) (1 pt), asymptotes \(y=\pm\dfrac{6}{8}x=\pm\dfrac{3}{4}x\). (1 pt)

5 Parametric equations / 4 pts

  1. \(\cos t=\dfrac{x-1}{4}\), \(\sin t=\dfrac{y+2}{3}\), so \(\dfrac{(x-1)^2}{16}+\dfrac{(y+2)^2}{9}=1\): an ellipse with center \((1,-2)\) (1 pt). At \(t=\dfrac{\pi}{2}\): \((1,1)\). (1 pt)
  2. \(t=\dfrac{x+1}{3}\), so \(y=\dfrac{(x+1)^2}{9}\), a parabola with vertex \((-1,0)\). (2 pts)
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