Hard Further Maths Conic sections 1 Questions

Challenging, exam-style Further Maths Conic sections 1 questions with worked solutions. Stretch yourself on the hardest conics, parabola, normal, intersection problems.

conicsparabolanormalintersectionhyperbolalocus
Further Maths34 questionsStep-by-step solutions
Question 1
8 markschallenging
The rectangular hyperbola HH has equation xy=36xy=36. The point P(6t, 6t)P\left(6t,\ \frac{6}{t}\right) lies on HH, where tt is a non-zero parameter. The tangents to HH at the points with parameters p=2p=2 and q=1q=1 meet at RR. Which of these is RR?
Show worked solution

Worked solution

  1. Write down the curve and its general parametric point

    xy=36,P(6t, 6t)xy=36,\quad P\left(6t,\ \frac{6}{t}\right)

    The parametric form generates every point of the conic as tt varies.

  2. Write the tangent at the point with parameter pp

    x+4y−24=0x + 4y - 24 = 0

    Use the standard tangent formula.

  3. Write the tangent at the point with parameter qq

    x+y−12=0x + y - 12 = 0

    Use the standard tangent formula again.

  4. Solve the pair of equations simultaneously

    x=8,y=4x=8,\quad y=4

    Eliminate one variable and back-substitute.

  5. Substitute the given parameter value

    p=2p=2

    Work with the specific point requested by the question.

  6. Write down the coordinates of the point of contact

    P(12, 3)P\left(12,\ 3\right)

    Substitute the parameter into the parametric form.

  7. Confirm the point lies on the curve

    12×3=3612\times 3=36

    Both sides agree, so the point really is on the conic.

  8. State the gradient of the tangent at this point

    dydx=−14\frac{dy}{dx}=- \frac{1}{4}

    This comes from differentiating the curve implicitly.

  9. State the gradient of the normal at this point

    mN=4m_{N}=4

    The normal gradient is the negative reciprocal of the tangent gradient.

  10. Record the value of the constant in the curve

    c=6c=6

    The constant fixes the size of the conic.

  11. Recall the standard parabola and its parametric point

    y2=4ax,P(at2, 2at)y^{2}=4ax,\quad P\left(at^{2},\ 2at\right)

    Every point of the parabola can be written in this parametric form.

  12. Recall the focus and directrix of y2=4axy^{2}=4ax

    S(a, 0),x=−aS\left(a,\ 0\right),\quad x=-a

    The focus is on the axis of symmetry and the directrix is the matching vertical line.

  13. Recall the standard tangent to the parabola

    ty=x+at2ty=x+at^{2}

    This is the tangent at the point with parameter tt.

  14. Recall the standard normal to the parabola

    y+tx=at3+2aty+tx=at^{3}+2at

    This is the normal at the point with parameter tt.

  15. Select the matching option

    R(8, 4)R\left(8,\ 4\right)

    This is the point of intersection.

Answer
(8, 4)\left(8,\ 4\right)
Question 2
8 markschallenging
The rectangular hyperbola HH has equation xy=16xy=16. The point P(4t, 4t)P\left(4t,\ \frac{4}{t}\right) lies on HH, where tt is a non-zero parameter. Which of these is an equation of the chord of HH joining the points with parameters p=2p=2 and q=−1q=-1?
Show worked solution

Worked solution

  1. Write down the curve and its general parametric point

    xy=16,P(4t, 4t)xy=16,\quad P\left(4t,\ \frac{4}{t}\right)

    The parametric form generates every point of the conic as tt varies.

  2. Find the two endpoints

    P(8, 2),Q(−4, −4)P\left(8,\ 2\right),\quad Q\left(-4,\ -4\right)

    Substitute each parameter into the parametric form.

  3. Find the gradient of the chord

    m=12m=\frac{1}{2}

    Use the change in yy over the change in xx.

  4. Use the point-gradient form through PP

    y−(2)=12(x−(8))y-\left(2\right)=\frac{1}{2}\left(x-\left(8\right)\right)

    Either endpoint may be used.

  5. Substitute the given parameter value

    p=2p=2

    Work with the specific point requested by the question.

  6. Write down the coordinates of the point of contact

    P(8, 2)P\left(8,\ 2\right)

    Substitute the parameter into the parametric form.

  7. Confirm the point lies on the curve

    8×2=168\times 2=16

    Both sides agree, so the point really is on the conic.

  8. State the gradient of the tangent at this point

    dydx=−14\frac{dy}{dx}=- \frac{1}{4}

    This comes from differentiating the curve implicitly.

  9. State the gradient of the normal at this point

    mN=4m_{N}=4

    The normal gradient is the negative reciprocal of the tangent gradient.

  10. Record the value of the constant in the curve

    c=4c=4

    The constant fixes the size of the conic.

  11. Recall the standard parabola and its parametric point

    y2=4ax,P(at2, 2at)y^{2}=4ax,\quad P\left(at^{2},\ 2at\right)

    Every point of the parabola can be written in this parametric form.

  12. Recall the focus and directrix of y2=4axy^{2}=4ax

    S(a, 0),x=−aS\left(a,\ 0\right),\quad x=-a

    The focus is on the axis of symmetry and the directrix is the matching vertical line.

  13. Recall the standard tangent to the parabola

    ty=x+at2ty=x+at^{2}

    This is the tangent at the point with parameter tt.

  14. Recall the standard normal to the parabola

    y+tx=at3+2aty+tx=at^{3}+2at

    This is the normal at the point with parameter tt.

  15. Select the matching option

    x−2y−4=0x - 2y - 4 = 0

    Clearing fractions gives this equation.

Answer
x−2y−4=0x - 2y - 4 = 0
Question 3
8 markschallenging
The rectangular hyperbola HH has equation xy=9xy=9. The point P(3t, 3t)P\left(3t,\ \frac{3}{t}\right) lies on HH, where tt is a non-zero parameter. The tangent to HH at PP meets the xx-axis at AA and the yy-axis at BB, and OO is the origin. Which of these is the area of triangle OABOAB?
Show worked solution

Worked solution

  1. Write down the curve and its general parametric point

    xy=9,P(3t, 3t)xy=9,\quad P\left(3t,\ \frac{3}{t}\right)

    The parametric form generates every point of the conic as tt varies.

  2. Differentiate the equation of the curve implicitly

    y+xdydx=0y+x\frac{dy}{dx}=0

    Apply the product rule to xyxy and differentiate the constant to zero.

  3. Rearrange to make the gradient the subject

    dydx=−yx\frac{dy}{dx}=-\frac{y}{x}

    Divide through by xx.

  4. Substitute the parametric coordinates

    dydx=−3t3t=−1t2\frac{dy}{dx}=-\frac{\frac{3}{t}}{3t}=-\frac{1}{t^{2}}

    The tangent gradient at the point with parameter tt is −1t2-\frac{1}{t^{2}}.

  5. Write the general tangent

    x+t2y=6tx+t^{2}y=6t

    This is the tangent at the point with parameter tt.

  6. Find the two axis intercepts

    A(6t, 0),B(0, 6t)A\left(6t,\ 0\right),\quad B\left(0,\ \frac{6}{t}\right)

    Set y=0y=0 and then x=0x=0 in the tangent.

  7. Compute the area of the right-angled triangle

    Area=12×6t×6t=18\text{Area}=\frac{1}{2}\times 6t\times\frac{6}{t}=18

    The parameter cancels, so the area is constant.

  8. Substitute the given parameter value

    t=1t=1

    Work with the specific point requested by the question.

  9. Write down the coordinates of the point of contact

    P(3, 3)P\left(3,\ 3\right)

    Substitute the parameter into the parametric form.

  10. Confirm the point lies on the curve

    3×3=93\times 3=9

    Both sides agree, so the point really is on the conic.

  11. State the gradient of the tangent at this point

    dydx=−1\frac{dy}{dx}=-1

    This comes from differentiating the curve implicitly.

  12. State the gradient of the normal at this point

    mN=1m_{N}=1

    The normal gradient is the negative reciprocal of the tangent gradient.

  13. Record the value of the constant in the curve

    c=3c=3

    The constant fixes the size of the conic.

  14. Recall the standard parabola and its parametric point

    y2=4ax,P(at2, 2at)y^{2}=4ax,\quad P\left(at^{2},\ 2at\right)

    Every point of the parabola can be written in this parametric form.

  15. Select the matching option

    Area=18\text{Area}=18

    The area is 2c22c^{2} for every point of the hyperbola.

Answer
1818
Question 4
8 markschallenging
The parabola CC has equation y2=8xy^{2}=8x. The point P(2t2, 4t)P\left(2t^{2},\ 4t\right) lies on CC, where tt is a parameter. The chord PQPQ of CC passes through the focus SS, and the point MM is the midpoint of PQPQ. Which of these is a Cartesian equation for the locus of MM as tt varies?
Show worked solution

Worked solution

  1. Write down the curve and its general parametric point

    y2=8x,P(2t2, 4t)y^{2}=8x,\quad P\left(2t^{2},\ 4t\right)

    The parametric form generates every point of the conic as tt varies.

  2. Write the coordinates of MM in terms of the parameter

    M(T4+1T2, 2T−2T)M\left(\frac{T^{4} + 1}{T^{2}},\ 2 T - \frac{2}{T}\right)

    Build the moving point from the description in the question.

  3. Write down the two parametric equations

    x=T4+1T2,y=2T−2Tx=\frac{T^{4} + 1}{T^{2}},\quad y=2 T - \frac{2}{T}

    These describe the locus parametrically.

  4. Eliminate the parameter

    remove t\text{remove } t

    Make the parameter the subject of one equation and substitute.

  5. Test a particular parameter value

    (174, 3)\left(\frac{17}{4},\ 3\right)

    This point must satisfy the correct Cartesian equation.

  6. Substitute the given parameter value

    t=2t=2

    Work with the specific point requested by the question.

  7. Write down the coordinates of the point of contact

    P(8, 8)P\left(8,\ 8\right)

    Substitute the parameter into the parametric form.

  8. Confirm the point lies on the curve

    (8)2=8×8\left(8\right)^{2}=8\times 8

    Both sides agree, so the point really is on the conic.

  9. State the gradient of the tangent at this point

    dydx=12\frac{dy}{dx}=\frac{1}{2}

    This comes from differentiating the curve implicitly.

  10. State the gradient of the normal at this point

    mN=−2m_{N}=-2

    The normal gradient is the negative reciprocal of the tangent gradient.

  11. Record the value of the constant in the curve

    a=2a=2

    The constant fixes the size of the conic.

  12. Recall the standard parabola and its parametric point

    y2=4ax,P(at2, 2at)y^{2}=4ax,\quad P\left(at^{2},\ 2at\right)

    Every point of the parabola can be written in this parametric form.

  13. Recall the focus and directrix of y2=4axy^{2}=4ax

    S(a, 0),x=−aS\left(a,\ 0\right),\quad x=-a

    The focus is on the axis of symmetry and the directrix is the matching vertical line.

  14. Recall the standard tangent to the parabola

    ty=x+at2ty=x+at^{2}

    This is the tangent at the point with parameter tt.

  15. Select the matching option

    y2=4x−8y^{2} = 4 x - 8

    This is the Cartesian equation of the locus.

Answer
y2=4x−8y^{2} = 4 x - 8
Question 5
8 markschallenging
The rectangular hyperbola HH has equation xy=16xy=16. The point P(4t, 4t)P\left(4t,\ \frac{4}{t}\right) lies on HH, where tt is a non-zero parameter. The tangents to HH at the points with parameters p=12p=\frac{1}{2} and q=13q=\frac{1}{3} meet at the point RR. Find the coordinates of RR.
Show worked solution

Worked solution

  1. Write down the curve and its general parametric point

    xy=16,P(4t, 4t)xy=16,\quad P\left(4t,\ \frac{4}{t}\right)

    The parametric form generates every point of the conic as tt varies.

  2. Write down the tangent at the point with parameter pp

    4x+y−16=04x + y - 16 = 0

    Use the standard tangent formula with t=pt=p.

  3. Write down the tangent at the point with parameter qq

    9x+y−24=09x + y - 24 = 0

    Use the standard tangent formula with t=qt=q.

  4. Solve the two tangent equations simultaneously

    {4x+y−16=09x+y−24=0\begin{cases}4x + y - 16 = 0\\ 9x + y - 24 = 0\end{cases}

    The intersection point satisfies both equations.

  5. Solve for the xx-coordinate

    x=85x=\frac{8}{5}

    Eliminate yy between the two equations.

  6. Solve for the yy-coordinate

    y=485y=\frac{48}{5}

    Back-substitute into either tangent.

  7. Substitute the given parameter value

    p=12p=\frac{1}{2}

    Work with the specific point requested by the question.

  8. Write down the coordinates of the point of contact

    P(2, 8)P\left(2,\ 8\right)

    Substitute the parameter into the parametric form.

  9. Confirm the point lies on the curve

    2×8=162\times 8=16

    Both sides agree, so the point really is on the conic.

  10. State the gradient of the tangent at this point

    dydx=−4\frac{dy}{dx}=-4

    This comes from differentiating the curve implicitly.

  11. State the gradient of the normal at this point

    mN=14m_{N}=\frac{1}{4}

    The normal gradient is the negative reciprocal of the tangent gradient.

  12. Record the value of the constant in the curve

    c=4c=4

    The constant fixes the size of the conic.

  13. Recall the standard parabola and its parametric point

    y2=4ax,P(at2, 2at)y^{2}=4ax,\quad P\left(at^{2},\ 2at\right)

    Every point of the parabola can be written in this parametric form.

  14. Recall the focus and directrix of y2=4axy^{2}=4ax

    S(a, 0),x=−aS\left(a,\ 0\right),\quad x=-a

    The focus is on the axis of symmetry and the directrix is the matching vertical line.

  15. State the coordinates of RR

    R(85, 485)R\left(\frac{8}{5},\ \frac{48}{5}\right)

    This is the point where the two tangents meet.

Answer
(85, 485)\left(\frac{8}{5},\ \frac{48}{5}\right)

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