Level 3 Differentiation Walkthrough

2016 NCEA Level 3 Differentiation Question 1(d)

2016 Paper

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Question

The tangents to the curve

\[y=\frac14(x-2)^2\]

at points P and Q are perpendicular.

Q is the point \((6,4)\).

Parabola with perpendicular tangents at P and Q The parabola y equals one quarter times x minus two squared has vertex at two comma zero. Q is at six comma four. Tangents at Q and at the point P on the left branch meet at a right angle. P Q (6, 4) x y

What is the \(x\)-coordinate of point P?

You must use calculus and show any derivatives that you need to find when solving this problem.

First walkthrough idea

Focus to try first

Find the gradient at Q, take its negative reciprocal, then use the derivative again to locate P.

Step 1

Differentiate the parabola

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

Show the first step’s working
\[ \frac{dy}{dx}=\frac14\cdot2(x-2)=\frac{x-2}{2}. \]

Walkthrough overview

What this question practises

This 2016 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Perpendicular tangent gradients on a parabola.

This is Question 1(d) from the 2016 NCEA Level 3 Differentiation paper for AS91578 — Apply differentiation methods in solving problems. Use the guided hints to practise the method before revealing the full working.

Calc.nz is an independent learning resource. Compare questions, diagrams, and assessment information with the official NZQA resources for AS91578.

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Learning summary

Review the method, not only the answer

This walkthrough helps you practise perpendicular tangent gradients on a parabola. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Use the derivative for the gradient and the original curve for the point before forming the tangent equation.

Continue practising

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