A-Level Tangents and normals Practice Questions

Free A-Level Tangents and normals practice questions with full step-by-step worked solutions. Covers equation of a tangent, differentiation, gradient at a point, cubic curves. Practise exam-style problems and check your method.

equation of a tangentdifferentiationgradient at a pointcubic curvesquadratic curvesequation of a normal
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
Find the equation of the tangent to the curve y=x2y = x^2 at the point where x=3x = 3.
Show worked solution

Worked solution

  1. Differentiate the curve to get the gradient function

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

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=3  ⇒  y=9x = 3 \;\Rightarrow\; y = 9

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=3=6\left.\frac{dy}{dx}\right|_{x=3} = 6

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Substitute the point and gradient

    y−9=6(x−3)y - 9 = 6(x - 3)

    Put the gradient and the point into the straight-line formula. We are almost there — just tidy it up.

  5. Simplify to the form y=mx+cy = mx + c

    y=6x−9y = 6 x - 9

    Expand the bracket and collect terms to write the tangent neatly. This is the equation of the tangent.

Answer
y=6x−9y = 6 x - 9
Question 2
3 markseasy
Find the equation of the normal to the curve y=x2+4xy = x^2 + 4x at the point where x=−1x = -1.
Show worked solution

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=2x+4\frac{dy}{dx} = 2 x + 4

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=−1  ⇒  y=−3x = -1 \;\Rightarrow\; y = -3

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=−1=2\left.\frac{dy}{dx}\right|_{x=-1} = 2

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Find the gradient of the normal

    mn=−12=−12m_n = -\dfrac{1}{2} = - \frac{1}{2}

    The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. Flip the fraction and change the sign.

  5. Substitute the point and normal gradient

    y+3=−12(x+1)y + 3 = - \frac{1}{2}(x + 1)

    Put the normal gradient and the point of contact into the straight-line formula.

  6. Simplify to the form y=mx+cy = mx + c

    y=−x2−72y = - \frac{x}{2} - \frac{7}{2}

    Expand and tidy up to get the normal in its neatest form. This is the equation of the normal.

Answer
y=−x2−72y = - \frac{x}{2} - \frac{7}{2}
Question 3
4 marksintermediate
Find the equation of the tangent to the curve y=x3−2x2y = x^3 - 2x^2 at the point where x=2x = 2.
Show worked solution

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=3x2−4x\frac{dy}{dx} = 3 x^{2} - 4 x

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=2  ⇒  y=0x = 2 \;\Rightarrow\; y = 0

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=2=4\left.\frac{dy}{dx}\right|_{x=2} = 4

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Recall the straight-line formula

    y−y1=m(x−x1)y - y_1 = m(x - x_1)

    A straight line through a known point uses this formula. Here mm is the gradient we just found and (x1,y1)(x_1,y_1) is the point of contact.

  5. Substitute the point and gradient

    y=4(x−2)y = 4(x - 2)

    Put the gradient and the point into the straight-line formula. We are almost there — just tidy it up.

  6. Simplify to the form y=mx+cy = mx + c

    y=4x−8y = 4 x - 8

    Expand the bracket and collect terms to write the tangent neatly. This is the equation of the tangent.

  7. Check the point lies on the line

    2→y=0=0 ✓2 \rightarrow y = 0 = 0\ \checkmark

    Substituting the xx-value back into the tangent should give the original yy-value. It matches, so the equation is correct.

Answer
y=4x−8y = 4 x - 8
Question 4
6 markshard
The tangent to the curve y=x2−4x+7y = x^2 - 4x + 7 at the point where x=3x = 3 meets the axes at AA and BB. Find the exact length ABAB.
Show worked solution

Worked solution

  1. Differentiate the curve

    dydx=2x−4\frac{dy}{dx} = 2 x - 4

    We need the gradient function to write the tangent's equation.

  2. Find the point of contact

    x=3⇒y=4x = 3 \Rightarrow y = 4

    Substitute the xx-value into the curve to find where the tangent touches.

  3. Find the tangent gradient

    dydx∣x=3=2\left.\frac{dy}{dx}\right|_{x=3} = 2

    This is the gradient of the tangent at the point of contact.

  4. Write the tangent equation

    y=2x−2y = 2 x - 2

    Use the straight-line formula with the gradient and the point, then tidy up.

  5. Find point AA on the xx-axis

    0=2x−2⇒A(1,0)0 = 2 x - 2 \Rightarrow A(1,0)

    Set y=0y = 0 to find where the tangent crosses the xx-axis.

  6. Find point BB on the yy-axis

    x=0⇒B(0,−2)x = 0 \Rightarrow B(0,-2)

    Set x=0x = 0 to find where the tangent crosses the yy-axis.

  7. Recall the distance formula

    AB=(x2−x1)2+(y2−y1)2AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

    The length of a line segment between two points uses Pythagoras. This is from the coordinate-geometry topic.

  8. Substitute the coordinates

    AB=(1)2+(−2)2AB = \sqrt{(1)^2 + (-2)^2}

    Put the coordinates of AA and BB into the distance formula.

  9. Work out the squares

    AB=1+4=5AB = \sqrt{1 + 4} = \sqrt{5}

    Square each coordinate difference and add them together under the square root.

  10. Simplify to an exact surd

    AB=5AB = \sqrt{5}

    Simplify the square root. Leaving it as a surd keeps the answer exact rather than rounding.

Answer
5\sqrt{5}
Question 5
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

Unlock 65 more Tangents and normals questions

Create a free account to work through every A-Level Tangents and normals question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Tangents and normals practice

Hard QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles