AS 91578 · 2023 paper · derivatives, tangents, a proof

Question 1(d) Merit

A chain-rule root, a product rate of change, a second tangent parallel to a given one, a parametric circle tangent to be proved in general form, and an optimisation whose answer is a fraction of a given area.

The diagram shows a tangent passing through the point P(p, q), which lies on the circle with parametric equations x = 4 cosθ and y = 4 sinθ.

Show that the equation of the tangent line is px + qy = p² + q².

x	4 cosθ
y	4 sinθ
P(p, q) on the circle

The 2023 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Differentiate x and y with respect to θ, and find dy/dx.
dx/dθ = −4 sinθ and dy/dθ = 4 cosθ
dy/dx = (4 cosθ)/(−4 sinθ)
2 At P, p = 4 cosθ and q = 4 sinθ. Write the gradient using p and q.
dy/dx = −(4 cosθ)/(4 sinθ) = −p/q
(Implicit differentiation of x² + y² = 16 gives the same thing, −x/y.)
3 Write the tangent through (p, q), and rearrange it.
y − q = −(p/q)(x − p)
Multiply by q: qy − q² = −px + p²
so px + qy = p² + q².

Answer The tangent at P is px + qy = p² + q², as required.

What the marker wanted

Achieved A correct dy/dx, written in θ, in x and y, or in p and q.
Merit The proof completed from a correct gradient.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

Jumping from the gradient to the result. NZQA’s 2023 report names this part among the proofs that many students could differentiate for but did not set out formally: show the line through (p, q), the multiplying by q, and the rearranging.

How did it go?

Got it moves it further out. Review brings it back tomorrow.