Further Maths Conic sections 2 Practice Questions

Free Further Maths Conic sections 2 practice questions with full step-by-step worked solutions. Covers conics, eccentricity, vertices, hyperbola. Practise exam-style problems and check your method.

conicseccentricityverticeshyperbolaasymptotesellipse
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The ellipse CC has equation x225+y29=1\frac{x^{2}}{25}+\frac{y^{2}}{9}=1. Find the eccentricity of CC.
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Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=25,b2=9a^{2}=25,\quad b^{2}=9

    The equation is already in the standard form for an ellipse.

  2. Quote the eccentricity relation for an ellipse

    b2=a2(1−e2)b^{2}=a^{2}\left(1-e^{2}\right)

    This is the relation given in the formula book.

  3. State the eccentricity of CC

    e=45e=\frac{4}{5}

    This is the eccentricity of the ellipse.

Answer
45\frac{4}{5}
Question 2
2 markseasy
The hyperbola CC has equation x21−y23=1\frac{x^{2}}{1}-\frac{y^{2}}{3}=1. Which of the following is the eccentricity of CC?
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Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=1,b2=3a^{2}=1,\quad b^{2}=3

    The equation is already in the standard form for a hyperbola.

  2. Quote the eccentricity relation for a hyperbola

    b2=a2(e2−1)b^{2}=a^{2}\left(e^{2}-1\right)

    This is the relation given in the formula book.

  3. Substitute the values of a2a^{2} and b2b^{2}

    3=1(e2−1)3=1\left(e^{2}-1\right)

    Both a2a^{2} and b2b^{2} are read straight off the equation.

  4. State the eccentricity of CC

    e=2e=2

    This is the eccentricity of the hyperbola.

Answer
22
Question 3
4 marksintermediate
The hyperbola CC has equation x216−y29=1\frac{x^{2}}{16}-\frac{y^{2}}{9}=1. Which of the following gives the equations of the directrices of CC?
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Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=16,b2=9a^{2}=16,\quad b^{2}=9

    The equation is already in the standard form for a hyperbola.

  2. Quote the eccentricity relation for a hyperbola

    b2=a2(e2−1)b^{2}=a^{2}\left(e^{2}-1\right)

    This is the relation given in the formula book.

  3. Substitute the values of a2a^{2} and b2b^{2}

    9=16(e2−1)9=16\left(e^{2}-1\right)

    Both a2a^{2} and b2b^{2} are read straight off the equation.

  4. Rearrange to make e2e^{2} the subject

    e2=2516e^{2}=\frac{25}{16}

    Divide by a2a^{2} and rearrange.

  5. Take the positive square root

    e=2516=54e=\sqrt{\frac{25}{16}}=\frac{5}{4}

    The eccentricity of a conic is positive.

  6. Write down the semi-axes

    a=4,b=3a=4,\quad b=3

    Take the positive square roots of a2a^{2} and b2b^{2}.

  7. State the equations of the directrices

    x=±165x=\pm \frac{16}{5}

    Both directrices are vertical lines.

Answer
x=±165x=\pm \frac{16}{5}
Question 4
6 markshard
The hyperbola CC has equation x236−y264=1\frac{x^{2}}{36}-\frac{y^{2}}{64}=1. Which of the following is an equation of the normal to CC at the point P(10,323)P\left(10,\frac{32}{3}\right)?
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Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=36,b2=64a^{2}=36,\quad b^{2}=64

    The equation is already in the standard form for a hyperbola.

  2. Check that PP lies on CC

    (10)236−(323)264=1\frac{\left(10\right)^{2}}{36}-\frac{\left(\frac{32}{3}\right)^{2}}{64}=1

    Substituting the coordinates of PP gives 11, so PP is on the curve.

  3. Differentiate the equation of CC implicitly with respect to xx

    2x36−2y64dydx=0\frac{2x}{36}-\frac{2y}{64}\frac{dy}{dx}=0

    Differentiate term by term, using the chain rule on the y2y^{2} term.

  4. Rearrange to make dydx\frac{dy}{dx} the subject

    dydx=16x9y\frac{dy}{dx}=\frac{16x}{9y}

    Collect the dydx\frac{dy}{dx} term and divide.

  5. Evaluate the gradient of the tangent at PP

    mT=53m_{T}=\frac{5}{3}

    Substitute the coordinates of PP into the derivative.

  6. Find the gradient of the normal

    mN=−1mT=−153=−35m_{N}=-\frac{1}{m_{T}}=-\frac{1}{\frac{5}{3}}=-\frac{3}{5}

    The normal is perpendicular to the tangent at PP.

  7. Write the normal in point-gradient form

    y−323=−35(x−10)y-\frac{32}{3}=-\frac{3}{5}\left(x-10\right)

    Use the gradient just found together with the coordinates of PP.

  8. Expand the right-hand side

    y=503−3x5y=\frac{50}{3}-\frac{3x}{5}

    Multiplying out gives yy explicitly in terms of xx.

  9. Verify that PP satisfies this equation

    9(10)+(15)(323)=2509\left(10\right)+\left(15\right)\left(\frac{32}{3}\right)=250

    The normal must pass through PP.

  10. State an equation of the normal

    9x+15y=2509x+15y=250

    This is the normal to CC at PP.

Answer
9x+15y=2509x+15y=250
Question 5
9 markschallenging
Which of the following is an equation of the hyperbola with foci (±17,0)\left(\pm 17,0\right) and eccentricity 1715\frac{17}{15}?
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Worked solution

  1. Use the foci to find aeae

    ae=17ae=17

    The foci of a conic in this standard position are at (±ae,0)\left(\pm ae,0\right).

  2. Divide by the eccentricity to find aa

    a=171715=15a=\frac{17}{\frac{17}{15}}=15

    Dividing aeae by ee leaves aa.

  3. Square to find a2a^{2}

    a2=225a^{2}=225

    This is the denominator of the x2x^{2} term.

  4. Quote the relation for b2b^{2}

    b2=a2e2−a2b^{2}=a^{2}e^{2}-a^{2}

    This follows from the eccentricity relation.

  5. Substitute a2a^{2} and aeae

    b2=172−225b^{2}=17^{2}-225

    Note a2e2=(ae)2a^{2}e^{2}=\left(ae\right)^{2}, which is known.

  6. Evaluate b2b^{2}

    b2=64b^{2}=64

    This is the denominator of the y2y^{2} term.

  7. Check the eccentricity of the conic found

    e=289225=1715e=\sqrt{\frac{289}{225}}=\frac{17}{15}

    The conic found does have the required eccentricity.

  8. Check the foci of the conic found

    ae=15×1715=17ae=15\times \frac{17}{15}=17

    The foci are in the required position.

  9. Recall the standard form of an ellipse

    x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

    Here aa is the semi-major axis and bb the semi-minor axis, with a>b>0a>b>0.

  10. Recall the standard form of a hyperbola

    x2a2−y2b2=1\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

    The hyperbola differs from the ellipse only in the sign of the y2y^{2} term.

  11. Recall the eccentricity relation for an ellipse

    b2=a2(1−e2)b^{2}=a^{2}\left(1-e^{2}\right)

    This links the two semi-axes to the eccentricity of an ellipse.

  12. Recall the eccentricity relation for a hyperbola

    b2=a2(e2−1)b^{2}=a^{2}\left(e^{2}-1\right)

    This is the hyperbola form of the same relation.

  13. Recall the position of the foci

    S(ae,0),S′(−ae,0)S\left(ae,0\right),\quad S'\left(-ae,0\right)

    Both foci lie on the xx-axis, symmetrically about the centre.

  14. Recall the equations of the directrices

    x=±aex=\pm\frac{a}{e}

    Each directrix is perpendicular to the axis through the foci.

  15. State an equation of the hyperbola

    x2225−y264=1\frac{x^{2}}{225}-\frac{y^{2}}{64}=1

    This is the required equation in standard form.

Answer
x2225−y264=1\frac{x^{2}}{225}-\frac{y^{2}}{64}=1

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