Hard Further Maths Conic sections 2 Questions

Challenging, exam-style Further Maths Conic sections 2 questions with worked solutions. Stretch yourself on the hardest conics, tangent, implicit-differentiation, normal problems.

conicstangentimplicit-differentiationnormaleccentricityforming-an-equation
Further Maths34 questionsStep-by-step solutions
Question 1
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}?
Show worked solution

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
Question 2
9 markschallenging
The ellipse CC has equation x2100+y236=1\frac{x^{2}}{100}+\frac{y^{2}}{36}=1. The normal to CC at the point P(6,245)P\left(6,\frac{24}{5}\right) meets the xx-axis at GG. Which of the following gives the coordinates of GG?
Show worked solution

Worked solution

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

    a2=100,b2=36a^{2}=100,\quad b^{2}=36

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

  2. Check that PP lies on CC

    (6)2100+(245)236=1\frac{\left(6\right)^{2}}{100}+\frac{\left(\frac{24}{5}\right)^{2}}{36}=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

    2x100+2y36dydx=0\frac{2x}{100}+\frac{2y}{36}\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=−9x25y\frac{dy}{dx}=-\frac{9x}{25y}

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

  5. Evaluate the gradient of the tangent at PP

    mT=−920m_{T}=-\frac{9}{20}

    Substitute the coordinates of PP into the derivative.

  6. Find the gradient of the normal

    mN=−1mT=209m_{N}=-\frac{1}{m_{T}}=\frac{20}{9}

    The normal is perpendicular to the tangent at PP.

  7. Write the normal in point-gradient form

    y−245=209(x−6)y-\frac{24}{5}=\frac{20}{9}\left(x-6\right)

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

  8. Expand the right-hand side

    y=20x9−12815y=\frac{20x}{9}-\frac{128}{15}

    Multiplying out gives yy explicitly in terms of xx.

  9. Clear the fractions and collect the terms

    100x−45y=384100x-45y=384

    Multiplying through by the common denominator gives integer coefficients.

  10. Verify that PP satisfies this equation

    100(6)+(−45)(245)=384100\left(6\right)+\left(-45\right)\left(\frac{24}{5}\right)=384

    The normal must pass through PP.

  11. Set y=0y=0 to find where the normal meets the xx-axis

    100x=384100x=384

    The xx-axis has equation y=0y=0.

  12. Solve for xx

    x=9625x=\frac{96}{25}

    Divide through by the coefficient of xx.

  13. Check that GG lies on the normal

    100(9625)+(−45)(0)=384100\left(\frac{96}{25}\right)+\left(-45\right)\left(0\right)=384

    Substituting GG back into the normal confirms the value.

  14. Note the standard result for an ellipse-type normal

    xG=(a2−b2)x1a2x_{G}=\frac{\left(a^{2}-b^{2}\right)x_{1}}{a^{2}}

    The xx-coordinate of GG is a fixed multiple of x1x_{1}.

  15. Confirm with that standard result

    xG=64×6100=9625x_{G}=\frac{64\times 6}{100}=\frac{96}{25}

    The two methods agree.

  16. 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.

  17. 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.

  18. State the coordinates of GG

    (9625,0)\left(\frac{96}{25},0\right)

    The normal meets the xx-axis at this point.

Answer
(9625,0)\left(\frac{96}{25},0\right)
Question 3
9 markschallenging
The hyperbola CC has equation x264−y236=1\frac{x^{2}}{64}-\frac{y^{2}}{36}=1. The tangent to CC at the point P(10,92)P\left(10,\frac{9}{2}\right) meets the xx-axis at AA and the yy-axis at BB. Which of the following is the area of triangle OABOAB, where OO is the origin?
Show worked solution

Worked solution

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

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

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

  2. Check that PP lies on CC

    (10)264−(92)236=1\frac{\left(10\right)^{2}}{64}-\frac{\left(\frac{9}{2}\right)^{2}}{36}=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

    2x64−2y36dydx=0\frac{2x}{64}-\frac{2y}{36}\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=9x16y\frac{dy}{dx}=\frac{9x}{16y}

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

  5. Evaluate the gradient of the tangent at PP

    mT=54m_{T}=\frac{5}{4}

    Substitute the coordinates of PP into the derivative.

  6. Write the tangent in point-gradient form

    y−92=54(x−10)y-\frac{9}{2}=\frac{5}{4}\left(x-10\right)

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

  7. Expand the right-hand side

    y=5x4−8y=\frac{5x}{4}-8

    Multiplying out gives yy explicitly in terms of xx.

  8. Clear the fractions and collect the terms

    5x−4y=325x-4y=32

    Multiplying through by the common denominator gives integer coefficients.

  9. Verify that PP satisfies this equation

    5(10)+(−4)(92)=325\left(10\right)+\left(-4\right)\left(\frac{9}{2}\right)=32

    The tangent must pass through PP.

  10. Make yy the subject of the tangent

    y=5x−324y=\frac{5x-32}{4}

    The tangent is rearranged ready to substitute into the curve.

  11. Substitute the tangent into the equation of CC and clear fractions

    −x236+5x9−259=0-\frac{x^{2}}{36}+\frac{5x}{9}-\frac{25}{9}=0

    This is the quadratic satisfied by the xx-coordinates of the intersections.

  12. Compute the discriminant of that quadratic

    Δ=(59)2−4(−136)(−259)=0\Delta=\left(\frac{5}{9}\right)^{2}-4\left(-\frac{1}{36}\right)\left(-\frac{25}{9}\right)=0

    A repeated root is expected if the line really is a tangent.

  13. Confirm the line touches CC exactly once

    Δ=0⇒repeated root\Delta=0\Rightarrow\text{repeated root}

    The zero discriminant proves the line is a tangent and not a chord.

  14. Find AA by setting y=0y=0 in the tangent

    5x=32⇒x=3255x=32\Rightarrow x=\frac{32}{5}

    The tangent crosses the xx-axis where y=0y=0.

  15. Write down the coordinates of AA

    (325,0)\left(\frac{32}{5},0\right)

    This is the point where the tangent meets the xx-axis.

  16. Find BB by setting x=0x=0 in the tangent

    −4y=32⇒y=−8-4y=32\Rightarrow y=-8

    The tangent crosses the yy-axis where x=0x=0.

  17. State the area of triangle OABOAB

    1285\frac{128}{5}

    This is the required area.

Answer
1285\frac{128}{5}
Question 4
9 markschallenging
The ellipse CC has equation x2100+y264=1\frac{x^{2}}{100}+\frac{y^{2}}{64}=1. The tangent to CC at the point P(6,325)P\left(6,\frac{32}{5}\right) meets the xx-axis at AA and the yy-axis at BB. Find the area of triangle OABOAB, where OO is the origin.
Show worked solution

Worked solution

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

    a2=100,b2=64a^{2}=100,\quad b^{2}=64

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

  2. Check that PP lies on CC

    (6)2100+(325)264=1\frac{\left(6\right)^{2}}{100}+\frac{\left(\frac{32}{5}\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

    2x100+2y64dydx=0\frac{2x}{100}+\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=−16x25y\frac{dy}{dx}=-\frac{16x}{25y}

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

  5. Evaluate the gradient of the tangent at PP

    mT=−35m_{T}=-\frac{3}{5}

    Substitute the coordinates of PP into the derivative.

  6. Write the tangent in point-gradient form

    y−325=−35(x−6)y-\frac{32}{5}=-\frac{3}{5}\left(x-6\right)

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

  7. Expand the right-hand side

    y=10−3x5y=10-\frac{3x}{5}

    Multiplying out gives yy explicitly in terms of xx.

  8. Clear the fractions and collect the terms

    3x+5y=503x+5y=50

    Multiplying through by the common denominator gives integer coefficients.

  9. Verify that PP satisfies this equation

    3(6)+(5)(325)=503\left(6\right)+\left(5\right)\left(\frac{32}{5}\right)=50

    The tangent must pass through PP.

  10. Make yy the subject of the tangent

    y=50−3x5y=\frac{50-3x}{5}

    The tangent is rearranged ready to substitute into the curve.

  11. Substitute the tangent into the equation of CC and clear fractions

    x264−3x16+916=0\frac{x^{2}}{64}-\frac{3x}{16}+\frac{9}{16}=0

    This is the quadratic satisfied by the xx-coordinates of the intersections.

  12. Compute the discriminant of that quadratic

    Δ=(−316)2−4(164)(916)=0\Delta=\left(-\frac{3}{16}\right)^{2}-4\left(\frac{1}{64}\right)\left(\frac{9}{16}\right)=0

    A repeated root is expected if the line really is a tangent.

  13. Confirm the line touches CC exactly once

    Δ=0⇒repeated root\Delta=0\Rightarrow\text{repeated root}

    The zero discriminant proves the line is a tangent and not a chord.

  14. Find AA by setting y=0y=0 in the tangent

    3x=50⇒x=5033x=50\Rightarrow x=\frac{50}{3}

    The tangent crosses the xx-axis where y=0y=0.

  15. Write down the coordinates of AA

    (503,0)\left(\frac{50}{3},0\right)

    This is the point where the tangent meets the xx-axis.

  16. State the area of triangle OABOAB

    2503\frac{250}{3}

    This is the required area.

Answer
2503\frac{250}{3}
Question 5
9 markschallenging
The ellipse CC has equation x2100+y264=1\frac{x^{2}}{100}+\frac{y^{2}}{64}=1. The normal to CC at the point P(8,245)P\left(8,\frac{24}{5}\right) meets the xx-axis at GG. Find the coordinates of GG.
Show worked solution

Worked solution

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

    a2=100,b2=64a^{2}=100,\quad b^{2}=64

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

  2. Check that PP lies on CC

    (8)2100+(245)264=1\frac{\left(8\right)^{2}}{100}+\frac{\left(\frac{24}{5}\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

    2x100+2y64dydx=0\frac{2x}{100}+\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=−16x25y\frac{dy}{dx}=-\frac{16x}{25y}

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

  5. Evaluate the gradient of the tangent at PP

    mT=−1615m_{T}=-\frac{16}{15}

    Substitute the coordinates of PP into the derivative.

  6. Find the gradient of the normal

    mN=−1mT=1516m_{N}=-\frac{1}{m_{T}}=\frac{15}{16}

    The normal is perpendicular to the tangent at PP.

  7. Write the normal in point-gradient form

    y−245=1516(x−8)y-\frac{24}{5}=\frac{15}{16}\left(x-8\right)

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

  8. Expand the right-hand side

    y=15x16−2710y=\frac{15x}{16}-\frac{27}{10}

    Multiplying out gives yy explicitly in terms of xx.

  9. Clear the fractions and collect the terms

    75x−80y=21675x-80y=216

    Multiplying through by the common denominator gives integer coefficients.

  10. Verify that PP satisfies this equation

    75(8)+(−80)(245)=21675\left(8\right)+\left(-80\right)\left(\frac{24}{5}\right)=216

    The normal must pass through PP.

  11. Set y=0y=0 to find where the normal meets the xx-axis

    75x=21675x=216

    The xx-axis has equation y=0y=0.

  12. Solve for xx

    x=7225x=\frac{72}{25}

    Divide through by the coefficient of xx.

  13. Check that GG lies on the normal

    75(7225)+(−80)(0)=21675\left(\frac{72}{25}\right)+\left(-80\right)\left(0\right)=216

    Substituting GG back into the normal confirms the value.

  14. Note the standard result for an ellipse-type normal

    xG=(a2−b2)x1a2x_{G}=\frac{\left(a^{2}-b^{2}\right)x_{1}}{a^{2}}

    The xx-coordinate of GG is a fixed multiple of x1x_{1}.

  15. State the coordinates of GG

    (7225,0)\left(\frac{72}{25},0\right)

    The normal meets the xx-axis at this point.

Answer
(7225,0)\left(\frac{72}{25},0\right)

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