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Date May 2018 Marks available 3 Reference code 18M.2.hl.TZ0.5
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 5 Adapted from N/A

Question

Consider the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).

The area enclosed by the ellipse is \(8\pi \) and \(b = 2\).

Show that the area enclosed by the ellipse is \(\pi ab\).

[9]
a.

Determine which coordinate axis the major axis of the ellipse lies along.

[2]
b.i.

Hence find the eccentricity.

[2]
b.ii.

Find the coordinates of the foci.

[2]
b.iii.

Find the equations of the directrices.

[2]
b.iv.

The centre of another ellipse is now given as the point (2, 1). The minor and major axes are of lengths 3 and 5 and are parallel to the \(x\) and \(y\) axes respectively. Find the equation of the ellipse.

[3]
c.

Markscheme

\(A = 4\int {y{\text{d}}x} \)      (M1)

\(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \Rightarrow \)

\(y = \frac{{b\sqrt {{a^2} - {x^2}} }}{a}\)      (A1)

let \(x = a\,{\text{cos}}\,\theta  \Rightarrow y = b\,{\text{sin}}\,\theta \)      M1

\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} =  - a\,{\text{sin}}\,\theta \)      A1

when \(x = 0,\,\,\theta  = \frac{\pi }{2}\). When \(x = a,\,\,\theta  = 0\)       A1

\( \Rightarrow A = 4\int_{\frac{\pi }{2}}^0 {b\,{\text{sin}}\,\theta } \left( { - a\,{\text{sin}}\,\theta } \right){\text{d}}\theta \)      M1

\( \Rightarrow A =  - 4ab\int_{\frac{\pi }{2}}^0 {\,{\text{si}}{{\text{n}}^2}\,\theta } \,{\text{d}}\theta \)

\( \Rightarrow A =  - 2ab\int_{\frac{\pi }{2}}^0 {\,\left( {1 - \,{\text{cos}}\,2\theta } \right)} \,{\text{d}}\theta \)      M1

\( \Rightarrow A =  - 2ab\left[ {\theta  - \frac{{{\text{sin}}\,2\theta }}{2}} \right]_{\frac{\pi }{2}}^0\)     A1

\( \Rightarrow A =  - 2ab\left[ {0 - 0 - \left( {\frac{\pi }{2} - 0} \right)} \right]\)     M1

\( \Rightarrow A = \pi ab\)      AG

[9 marks]

a.

\(b = 2\)

hence \(2\pi a = 8\pi  \Rightarrow a = 4\)      A1

hence major axis lies along the x-axis      A1

[2 marks]

b.i.

\({b^2} = {a^2}\left( {1 - {e^2}} \right)\)      (M1)

\(4 = 16\left( {1 - {e^2}} \right) \Rightarrow e = \frac{{\sqrt 3 }}{2}\)      A1

[2 marks]

b.ii.

coordinates of foci are \(\left( { \pm ae,\,0} \right) = \left( {2\sqrt 3 ,\,0} \right),\,\left( { - 2\sqrt 3 ,\,0} \right)\)      A1A1

[2 marks]

b.iii.

equations of directrices are \(x =  \pm \frac{a}{e} = \frac{8}{{\sqrt 3 }},\, - \frac{8}{{\sqrt 3 }}\)      A1A1

[2 marks]

b.iv.

\(a = \frac{3}{2},\,b = \frac{5}{2}\)     (A1)

hence equation is \(\frac{4}{9}{\left( {x - 2} \right)^2} + \frac{4}{{25}}{\left( {y - 1} \right)^2} = 1\)     M1A1

[3 marks]

c.

Examiners report

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b.i.
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b.ii.
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b.iii.
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b.iv.
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c.

Syllabus sections

Topic 2 - Geometry » 2.6 » Conic sections.

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