The ellipse and hyperbola

The ellipse and hyperbola Quiz

Spec 2.1 · Edexcel International A Level Further Maths: Quiz

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5 questions, one answer each

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  1. 1. The Cartesian equation of x=2cos⁡tx=2\cos t, y=5sin⁡ty=5\sin t is

    • A. x225+y24=1\frac{x^2}{25}+\frac{y^2}{4}=1
    • B. x24+y225=1\frac{x^2}{4}+\frac{y^2}{25}=1
    • C. x22+y25=1\frac{x^2}{2}+\frac{y^2}{5}=1
    • D. 2x2+5y2=12x^2+5y^2=1
  2. 2. Which parametrisation traces only the right branch of x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1?

    • A. x=acos⁡tx=a\cos t, y=bsin⁡ty=b\sin t
    • B. x=asec⁡tx=a\sec t, y=btan⁡ty=b\tan t for all tt
    • C. x=asinh⁡tx=a\sinh t, y=bcosh⁡ty=b\cosh t
    • D. x=acosh⁡tx=a\cosh t, y=bsinh⁡ty=b\sinh t
  3. 3. To derive the Cartesian equation from x=asec⁡tx=a\sec t, y=btan⁡ty=b\tan t you use

    • A. 1+tan⁡2t=sec⁡2t1+\tan^2t=\sec^2t
    • B. sin⁡2t+cos⁡2t=1\sin^2t+\cos^2t=1
    • C. cosh⁡2t−sinh⁡2t=1\cosh^2t-\sinh^2t=1
    • D. cos⁡2t=1−2sin⁡2t\cos2t=1-2\sin^2t
  4. 4. For the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1, the semi-major axis has length

    • A. 22
    • B. 44
    • C. 33
    • D. 99
  5. 5. Which parametrisation gives the left branch of x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1?

    • A. x=acosh⁡tx=a\cosh t, y=−bsinh⁡ty=-b\sinh t
    • B. x=−acosh⁡tx=-a\cosh t, y=bsinh⁡ty=b\sinh t
    • C. x=asinh⁡tx=a\sinh t, y=bcosh⁡ty=b\cosh t
    • D. x=asech⁡tx=a\operatorname{sech}t, y=btanh⁡ty=b\tanh t

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