Hard A-Level Tangents and normals Questions

Challenging, exam-style A-Level Tangents and normals questions with worked solutions. Stretch yourself on the hardest equation of a tangent, intersection of line and curve, area of a triangle, equation of a normal problems.

equation of a tangentintersection of line and curvearea of a triangleequation of a normaltangents from a pointcubic curves
A-Level34 questionsStep-by-step solutions
Question 1
6 markschallenging
Find the values of kk for which the line y=kx−4y = kx - 4 is a tangent to the curve y=x2y = x^2.
Show worked solution

Worked solution

  1. State the tangency idea

    tangent  ⟺  one repeated intersection\text{tangent} \iff \text{one repeated intersection}

    A line is a tangent when it touches the curve at exactly one point, i.e. the intersection equation has a repeated root.

  2. Set the line equal to the curve

    x2=kx−4x^{2} = k x - 4

    Where the line meets the curve their yy-values agree, so we set the expressions equal.

  3. Rearrange into a quadratic in xx

    −kx+x2+4=0- k x + x^{2} + 4 = 0

    Bring everything to one side. The coefficients now contain the unknown gradient kk.

  4. Identify the coefficients

    a=1, b=−k, c=4a = 1,\ b = - k,\ c = 4

    Compare with ax2+bx+c=0ax^2+bx+c=0 ready to use the discriminant.

  5. Recall the discriminant condition

    b2−4ac=0b^2 - 4ac = 0

    One repeated root means the discriminant is zero. This is the tangency test.

  6. Compute b2b^2

    b2=k2b^2 = k^{2}

    Square the coefficient of xx.

  7. Compute 4ac4ac

    4ac=164ac = 16

    Multiply four, aa and cc together.

  8. Form the discriminant equation

    k2−16=0k^{2} - 16 = 0

    Subtract and set to zero. This is a quadratic in kk, so expect two answers.

  9. Solve for kk

    k=−4, 4k = -4,\ 4

    Solving gives the two gradients that make the line touch the curve.

  10. Interpret the two solutions

    two tangent lines with these gradients\text{two tangent lines with these gradients}

    Each gradient gives a distinct tangent line to the curve, which is why there are two answers.

  11. State the values of kk

    k=−4, 4k = -4,\ 4

    These are the required gradient values.

  12. Check the answer is reasonable

    result: k=−4, 4\text{result}:\ k = -4,\ 4

    We re-read the question and confirm this result answers exactly what was asked, with a sensible size and sign.

  13. Summarise the method

    differentiate→gradient→straight line\text{differentiate} \to \text{gradient} \to \text{straight line}

    The key routine of this topic is to differentiate for a gradient, then use the straight-line formula. Keeping this order avoids mistakes.

  14. Watch the common slip

    mnormal=−1mtangentm_{\text{normal}} = -\tfrac{1}{m_{\text{tangent}}}

    Many students confuse tangent and normal gradients. Remember the normal's gradient is the negative reciprocal of the tangent's.

  15. Keep the answer exact

    k=−4, 4k = -4,\ 4

    We leave the answer as an exact fraction or surd rather than a rounded decimal, which is what A-Level expects.

Answer
k=−4, 4k = -4,\ 4
Question 2
6 markschallenging
Explain why the curve y=x3y = x^3 has exactly one point at which the tangent is horizontal, and state the coordinates of that point. Select the fully correct explanation.
Show worked solution

Worked solution

  1. State what a horizontal tangent needs

    dydx=0\frac{dy}{dx} = 0

    A tangent is horizontal exactly where the gradient of the curve is zero.

  2. Differentiate the curve

    dydx=3x2\frac{dy}{dx} = 3x^2

    The gradient function of y=x3y = x^3 is 3x23x^2.

  3. Set the gradient to zero

    3x2=03x^2 = 0

    Horizontal tangents occur where this equals zero.

  4. Solve the equation

    x2=0⇒x=0x^2 = 0 \Rightarrow x = 0

    Dividing by 33 gives x2=0x^2 = 0, whose only solution is x=0x = 0.

  5. Count the solutions

    x=0 only (repeated)x = 0 \text{ only (repeated)}

    Because x2=0x^2 = 0 has just one (repeated) root, there is only one such xx-value.

  6. Explain why only one

    3x2≥0 and =0 only at 03x^2 \geq 0 \text{ and } = 0 \text{ only at } 0

    The expression 3x23x^2 is never negative and equals zero only at x=0x = 0, so no other point can have a horizontal tangent.

  7. Find the yy-coordinate

    y=03=0y = 0^3 = 0

    Substitute x=0x = 0 into the curve to get the point.

  8. State the point

    (0,0)(0, 0)

    The single horizontal tangent occurs at the origin.

  9. Note the shape there

    point of inflection, not a turning point\text{point of inflection, not a turning point}

    At x=0x = 0 the gradient touches zero but does not change sign, so it is a point of inflection rather than a maximum or minimum.

  10. Check the sign of the gradient

    3x2>0 for x≠03x^2 > 0 \text{ for } x \neq 0

    Either side of the origin the gradient is positive, confirming the curve keeps increasing.

  11. Conclude the explanation

    exactly one horizontal tangent at (0,0)\text{exactly one horizontal tangent at } (0,0)

    Since 3x2=03x^2 = 0 has the single root x=0x = 0, there is exactly one horizontal tangent, at the origin.

  12. Check the answer is reasonable

    \text{result}:\ Since dydx=3x2=0\tfrac{dy}{dx} = 3x^2 = 0 only at x=0x = 0 (and 3x2≥03x^2 \geq 0 elsewhere), there is exactly one horizontal tangent, at (0,0)(0, 0).

    We re-read the question and confirm this result answers exactly what was asked, with a sensible size and sign.

  13. Summarise the method

    differentiate→gradient→straight line\text{differentiate} \to \text{gradient} \to \text{straight line}

    The key routine of this topic is to differentiate for a gradient, then use the straight-line formula. Keeping this order avoids mistakes.

  14. Watch the common slip

    mnormal=−1mtangentm_{\text{normal}} = -\tfrac{1}{m_{\text{tangent}}}

    Many students confuse tangent and normal gradients. Remember the normal's gradient is the negative reciprocal of the tangent's.

  15. Keep the answer exact

    Since dydx=3x2=0\tfrac{dy}{dx} = 3x^2 = 0 only at x=0x = 0 (and 3x2≥03x^2 \geq 0 elsewhere), there is exactly one horizontal tangent, at (0,0)(0, 0).

    We leave the answer as an exact fraction or surd rather than a rounded decimal, which is what A-Level expects.

Answer
Since dydx=3x2=0\tfrac{dy}{dx} = 3x^2 = 0 only at x=0x = 0 (and 3x2≥03x^2 \geq 0 elsewhere), there is exactly one horizontal tangent, at (0,0)(0, 0).
Question 3
6 markschallenging
Find the coordinates of the point where the tangents to the curve y=x2y = x^2 at the points where x=1x = 1 and x=3x = 3 meet.
Show worked solution

Worked solution

  1. Differentiate the curve

    dydx=2x\frac{dy}{dx} = 2 x

    The gradient function gives the tangent gradient at any point on the curve.

  2. Point and gradient at x=1x = 1

    (1,1),m1=2(1,1),\quad m_1 = 2

    Find the point on the curve and the gradient there. This is the first point of contact.

  3. Write the first tangent

    y=2x−1y = 2 x - 1

    Use the straight-line formula for the first tangent.

  4. Point and gradient at x=3x = 3

    (3,9),m2=6(3,9),\quad m_2 = 6

    Repeat for the second point: find the point on the curve and the gradient there.

  5. Write the second tangent

    y=6x−9y = 6 x - 9

    Use the straight-line formula for the second tangent.

  6. Set the two tangents equal

    2x−1=6x−92 x - 1 = 6 x - 9

    The tangents meet where their yy-values are equal. Setting the equations equal finds that meeting point.

  7. Solve for xx

    x=2x = 2

    Collect the xx-terms and solve. This is the xx-coordinate of the intersection.

  8. Find the yy-coordinate

    y=3y = 3

    Substitute the xx-value into either tangent to get the yy-coordinate.

  9. State the meeting point

    (2, 3)\left(2,\ 3\right)

    This is where the two tangents cross each other.

  10. Check the answer is reasonable

    result: (2, 3)\text{result}:\ \left(2,\ 3\right)

    We re-read the question and confirm this result answers exactly what was asked, with a sensible size and sign.

  11. Summarise the method

    differentiate→gradient→straight line\text{differentiate} \to \text{gradient} \to \text{straight line}

    The key routine of this topic is to differentiate for a gradient, then use the straight-line formula. Keeping this order avoids mistakes.

  12. Watch the common slip

    mnormal=−1mtangentm_{\text{normal}} = -\tfrac{1}{m_{\text{tangent}}}

    Many students confuse tangent and normal gradients. Remember the normal's gradient is the negative reciprocal of the tangent's.

  13. Keep the answer exact

    (2, 3)\left(2,\ 3\right)

    We leave the answer as an exact fraction or surd rather than a rounded decimal, which is what A-Level expects.

  14. Relate the answer to the graph

    the line meets the curve as expected\text{the line meets the curve as expected}

    Picturing the curve and line together is a quick way to confirm the answer makes geometric sense.

  15. State the final conclusion

    (2, 3)\left(2,\ 3\right)

    We present the final answer clearly and neatly so it is easy to read and to mark.

Answer
(2, 3)\left(2,\ 3\right)
Question 4
6 markschallenging
The tangents to the curve y=x2y = x^2 at the points where x=−1x = -1 and x=2x = 2 meet at a point PP. Find the area of the triangle formed by PP and the two points of contact.
Show worked solution

Worked solution

  1. Differentiate the curve

    dydx=2x\frac{dy}{dx} = 2 x

    The gradient function gives the tangent gradient at any point.

  2. Find the tangency point AA at x=−1x = -1

    A=(−1,1), m1=−2A = (-1,1),\ m_1 = -2

    Find the point on the curve and the gradient of the tangent there.

  3. Write the first tangent

    y=−2x−1y = - 2 x - 1

    Use the straight-line formula for the tangent at AA.

  4. Find the tangency point BB at x=2x = 2

    B=(2,4), m2=4B = (2,4),\ m_2 = 4

    Find the point on the curve and the gradient of the tangent there.

  5. Write the second tangent

    y=4x−4y = 4 x - 4

    Use the straight-line formula for the tangent at BB.

  6. Set the tangents equal

    −2x−1=4x−4- 2 x - 1 = 4 x - 4

    The two tangents meet where their equations agree.

  7. Solve for the meeting xx

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

    Collecting terms and solving gives the xx-coordinate of the meeting point PP.

  8. Find the meeting yy

    y=−2y = -2

    Substitute back to get the yy-coordinate of PP.

  9. List the three vertices

    A(−1,1), B(2,4), P(12,−2)A(-1,1),\ B(2,4),\ P(\frac{1}{2},-2)

    The triangle has the two tangency points and the meeting point as its corners.

  10. Recall the triangle-area formula

    Area=12∣xA(yB−yP)+xB(yP−yA)+xP(yA−yB)∣\text{Area} = \tfrac12\left|x_A(y_B-y_P)+x_B(y_P-y_A)+x_P(y_A-y_B)\right|

    This shoelace formula finds the area of a triangle from its three vertices. It always gives a positive value.

  11. Substitute the coordinates

    Area=12∣−1(4−−2)+2(−2−1)+12(1−4)∣\text{Area} = \tfrac12\left|-1(4--2)+2(-2-1)+\frac{1}{2}(1-4)\right|

    Carefully put each coordinate into the formula, keeping track of signs.

  12. Evaluate the area

    Area=274\text{Area} = \frac{27}{4}

    Working through the arithmetic gives the exact area of the triangle.

  13. Check the answer is reasonable

    result: 274\text{result}:\ \frac{27}{4}

    We re-read the question and confirm this result answers exactly what was asked, with a sensible size and sign.

  14. Summarise the method

    differentiate→gradient→straight line\text{differentiate} \to \text{gradient} \to \text{straight line}

    The key routine of this topic is to differentiate for a gradient, then use the straight-line formula. Keeping this order avoids mistakes.

  15. Watch the common slip

    mnormal=−1mtangentm_{\text{normal}} = -\tfrac{1}{m_{\text{tangent}}}

    Many students confuse tangent and normal gradients. Remember the normal's gradient is the negative reciprocal of the tangent's.

Answer
274\frac{27}{4}
Question 5
6 markschallenging
The tangent to the curve y=x3−2xy = x^3 - 2x at the point where x=1x = 1 meets the curve again. Find the coordinates of the second point.
Show worked solution

Worked solution

  1. Differentiate the curve

    dydx=3x2−2\frac{dy}{dx} = 3 x^{2} - 2

    We first need the gradient function so we can build the equation of the line.

  2. Find the point of contact

    x=1⇒y=−1x = 1 \Rightarrow y = -1

    Substitute the given xx-value into the curve to find the starting point.

  3. Find the tangent gradient

    dydx∣x=1=1\left.\frac{dy}{dx}\right|_{x=1} = 1

    This is the gradient of the tangent where the line touches the curve.

  4. Recall the straight-line formula

    y−y1=m(x−x1)y - y_1 = m(x - x_1)

    This lets us write the line through the known point with the gradient we found.

  5. Write the equation of the tangent

    y=x−2y = x - 2

    Substitute the gradient and the point into the straight-line formula and simplify.

  6. Set the curve equal to the line

    x3−2x=x−2x^{3} - 2 x = x - 2

    Where the line meets the curve, their yy-values are equal. Setting them equal finds all intersection points.

  7. Bring everything to one side

    x3−3x+2=0x^{3} - 3 x + 2 = 0

    Rearranging into a single polynomial equal to zero lets us solve for the xx-coordinates of the intersections.

  8. Note the structure of the equation

    deg⁡=3\deg = 3

    The highest power tells us how many intersection points to expect at most. We already know one of them is the point of contact.

  9. Factorise using the repeated root

    (x−1)2(x+2)=0\left(x - 1\right)^{2} \left(x + 2\right) = 0

    Because the line is a tangent, x=1x = 1 is a repeated (double) root. That factor appears twice, which makes factorising easier.

  10. List all the roots

    x=−2, 1x = -2,\ 1

    The solutions include the original point of contact. Any solution different from it is where the line meets the curve again.

  11. Select the second intersection

    x=−2x = -2

    We discard x=1x = 1 (the original point) and keep the new value.

  12. Find its yy-coordinate

    x=−2⇒y=−4x = -2 \Rightarrow y = -4

    Substitute the new xx-value into the curve (or the line) to find the matching yy-value.

  13. State the coordinates of the second point

    (−2, −4)\left(-2,\ -4\right)

    This is the point where the line meets the curve for a second time.

  14. Check by substituting into the line

    y=−4=−4 ✓y = -4 = -4\ \checkmark

    The point should also lie on the line, since it is an intersection. It matches, confirming the answer.

  15. Interpret the result

    line meets curve again at (−2, −4)\text{line meets curve again at }\left(-2,\ -4\right)

    Geometrically, the line leaves the point of contact and cuts the curve once more at this second point. This is exactly what the question asked for.

Answer
(−2, −4)\left(-2,\ -4\right)

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