Question
Consider the ellipse having equation \({x^2} + 3{y^2} = 2\).
(i) Find the equation of the tangent to the ellipse at the point \(\left( {1,{\text{ }}\frac{1}{{\sqrt 3 }}} \right)\).
(ii) Find the equation of the normal to the ellipse at the point \(\left( {1,{\text{ }}\frac{1}{{\sqrt 3 }}} \right)\).
Given that the tangent crosses the \(x\)-axis at P and the normal crosses the \(y\)-axis at Q, find the equation of (PQ).
Hence show that (PQ) touches the ellipse.
State the coordinates of the point where (PQ) touches the ellipse.
Find the coordinates of the foci of the ellipse.
Find the equations of the directrices of the ellipse.
Answer/Explanation
Markscheme
(i) \(2x + 6y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{x}{{3y}}\) A1
gradient of tangent is \( – \frac{{\sqrt 3 }}{3}\) A1
equation of tangent is \(y – \frac{1}{{\sqrt 3 }} = – \frac{{\sqrt 3 }}{3}(x – 1)\) M1A1
\(\left( { \Rightarrow y = – \frac{{\sqrt 3 }}{3}x + \frac{{\sqrt 3 }}{3} + \frac{1}{{\sqrt 3 }} \Rightarrow y = – \frac{{\sqrt 3 }}{3}x + \frac{2}{{\sqrt 3 }}} \right)\)
(ii) gradient of normal is \(\sqrt 3 \) A1
equation of normal is \(y – \frac{1}{{\sqrt 3 }} = \sqrt 3 (x – 1)\) A1
\(\left( { \Rightarrow y = x\sqrt 3 – \sqrt 3 + \frac{1}{{\sqrt 3 }} \Rightarrow y = \sqrt 3 x – \frac{2}{{\sqrt 3 }}} \right)\)
coordinates of P are \((2,{\text{ }}0)\) A1
coordinates of Q are \(\left( {0,{\text{ }} – \frac{2}{{\sqrt 3 }}} \right)\) A1
equation of (PQ) is \(\frac{{y – 0}}{{x – 2}} = \frac{{\frac{2}{{\sqrt 3 }}}}{2}\) M1
\( \Rightarrow y = \frac{1}{{\sqrt 3 }}(x – 2)\) A1
substitute equation of (PQ) into equation of ellipse
\({x^2} + 3{\left( {\frac{{x – 2}}{{\sqrt 3 }}} \right)^2} = 2\) M1A1
\( \Rightarrow {x^2} + {x^2} – 4x + 4 = 2\)
\( \Rightarrow {(x – 1)^2} = 0\) A1
since the equation has two equal roots (PQ) touches the ellipse R1
\(\left( {1,{\text{ }} – \frac{1}{{\sqrt 3 }}} \right)\) A1
\({x^2} + 3{y^2} = 2\)
\(\frac{{{x^2}}}{2} + \frac{{{y^2}}}{{\frac{2}{3}}} = 1\)
\( \Rightarrow a = \sqrt 2 ,{\text{ }}b = \sqrt {\frac{2}{3}} \) A1
EITHER
\({b^2} = {a^2}(1 – {e^2})\)
\(\frac{2}{3} = 2(1 – {e^2})\) M1
\( \Rightarrow e = \sqrt {\frac{2}{3}} \) A1
coordinates of foci are \(( \pm ae,{\text{ }}0) \Rightarrow \left( {\frac{2}{{\sqrt 3 }},{\text{ }}0} \right),{\text{ }}\left( { – \frac{2}{{\sqrt 3 }},{\text{ }}0} \right)\) A1
OR
\({f^2} = {a^2} – {b^2}\) M1
\({f^2} = 4 – \frac{2}{3}\) A1
coordinates of foci are \(\left( {\frac{2}{{\sqrt 3 }},{\text{ }}0} \right),{\text{ }}\left( { – \frac{2}{{\sqrt 3 }},{\text{ }}0} \right)\) A1
Note: Award accuracy marks if \({a^2}\), \({b^2}\) and \({e^2}\) are given.
EITHER
equations of directrices are \(x = \pm \frac{a}{e} \Rightarrow x = \sqrt 3 ,{\text{ }}x = – \sqrt 3 \) A1
OR
\(d = \frac{{{a^2}}}{f} \Rightarrow x = \sqrt 3 ,{\text{ }}x = – \sqrt 3 \) A1
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 \(\pi ab\).
Determine which coordinate axis the major axis of the ellipse lies along.
Hence find the eccentricity.
Find the coordinates of the foci.
Find the equations of the directrices.
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.
Answer/Explanation
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]
\(b = 2\)
hence \(2\pi a = 8\pi \Rightarrow a = 4\) A1
hence major axis lies along the x-axis A1
[2 marks]
\({b^2} = {a^2}\left( {1 – {e^2}} \right)\) (M1)
\(4 = 16\left( {1 – {e^2}} \right) \Rightarrow e = \frac{{\sqrt 3 }}{2}\) A1
[2 marks]
coordinates of foci are \(\left( { \pm ae,\,0} \right) = \left( {2\sqrt 3 ,\,0} \right),\,\left( { – 2\sqrt 3 ,\,0} \right)\) A1A1
[2 marks]
equations of directrices are \(x = \pm \frac{a}{e} = \frac{8}{{\sqrt 3 }},\, – \frac{8}{{\sqrt 3 }}\) A1A1
[2 marks]
\(a = \frac{3}{2},\,b = \frac{5}{2}\) (A1)
hence equation is \(\frac{4}{9}{\left( {x – 2} \right)^2} + \frac{4}{{25}}{\left( {y – 1} \right)^2} = 1\) M1A1
[3 marks]
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