Further Maths Conic sections 1 Practice Questions

Free Further Maths Conic sections 1 practice questions with full step-by-step worked solutions. Covers conics, parabola, parametric-form, hyperbola. Practise exam-style problems and check your method.

conicsparabolaparametric-formhyperbolafocusdirectrix
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The parabola CC has equation y2=12xy^{2}=12x. The point P(3t2, 6t)P\left(3t^{2},\ 6t\right) lies on CC, where tt is a parameter. Find the coordinates of PP when t=2t=2.
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Worked solution

  1. Write down the curve and its general parametric point

    y2=12x,P(3t2, 6t)y^{2}=12x,\quad P\left(3t^{2},\ 6t\right)

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

  2. Evaluate the yy-coordinate

    y=12y=12

    Work out the second parametric coordinate.

  3. State the coordinates of PP

    P(12, 12)P\left(12,\ 12\right)

    These are the required coordinates.

Answer
(12, 12)\left(12,\ 12\right)
Question 2
2 markseasy
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. Which of these is the gradient of the normal to CC at the point with parameter t=3t=3?
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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. Take the negative reciprocal

    mnormal=−113=−3m_{\text{normal}}=-\frac{1}{\frac{1}{3}}=-3

    The normal is perpendicular to the tangent.

  3. Select the matching option

    mN=−3m_{N}=-3

    This is the required normal gradient.

Answer
−3-3
Question 3
4 marksintermediate
The rectangular hyperbola HH has equation xy=25xy=25. The point P(5t, 5t)P\left(5t,\ \frac{5}{t}\right) lies on HH, where tt is a non-zero parameter. Which of these is the point on HH with parameter t=12t=\frac{1}{2}?
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Worked solution

  1. Write down the curve and its general parametric point

    xy=25,P(5t, 5t)xy=25,\quad P\left(5t,\ \frac{5}{t}\right)

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

  2. Substitute the given parameter value

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

    Replace tt by the value stated in the question.

  3. Evaluate both parametric coordinates

    x=52,y=10x=\frac{5}{2},\quad y=10

    Work out each coordinate separately.

  4. Substitute the given parameter value

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

    Work with the specific point requested by the question.

  5. Confirm the point lies on the curve

    52×10=25\frac{5}{2}\times 10=25

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

  6. Select the matching option

    P(52, 10)P\left(\frac{5}{2},\ 10\right)

    This option matches the computed coordinates.

Answer
(52, 10)\left(\frac{5}{2},\ 10\right)
Question 4
6 markshard
The rectangular hyperbola HH has equation xy=4xy=4. The point P(2t, 2t)P\left(2t,\ \frac{2}{t}\right) lies on HH, where tt is a non-zero parameter. Which of these is an equation of the normal to HH at the point with parameter t=2t=2?
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Worked solution

  1. Write down the curve and its general parametric point

    xy=4,P(2t, 2t)xy=4,\quad P\left(2t,\ \frac{2}{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=−2t2t=−1t2\frac{dy}{dx}=-\frac{\frac{2}{t}}{2t}=-\frac{1}{t^{2}}

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

  5. Find the coordinates of the point of contact

    P(4, 1)P\left(4,\ 1\right)

    Substitute the given parameter value into the parametric form.

  6. Find the normal gradient

    mnormal=−1−14=4m_{\text{normal}}=-\frac{1}{- \frac{1}{4}}=4

    Take the negative reciprocal of the tangent gradient.

  7. Use the point-gradient form

    y−(1)=4(x−(4))y-\left(1\right)=4\left(x-\left(4\right)\right)

    Combine the normal gradient with the point of contact.

  8. Substitute the given parameter value

    t=2t=2

    Work with the specific point requested by the question.

  9. Write down the coordinates of the point of contact

    P(4, 1)P\left(4,\ 1\right)

    Substitute the parameter into the parametric form.

  10. Select the matching option

    4x−y−15=04x - y - 15 = 0

    Clearing fractions gives this equation.

Answer
4x−y−15=04x - y - 15 = 0
Question 5
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?
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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)

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