NCEA Calculus 2022 TWO(e - Solution & Explanation

Question

The curve with the equation

(y-5)^2 = 16(x-2)

has a tangent of gradient 1 at point P.

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This tangent intersects the

x

and

y

axes at points R and S respectively.

Prove that the length RS is

7\sqrt{2}

.

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

Official Answer

(y-5)^2=16(x-2)

Method A

y-5=4\sqrt{x-2}

y=4\sqrt{x-2}+5

\frac{dy}{dx}=\frac{2}{\sqrt{x-2}}

\frac{dy}{dx}=1

\frac{2}{\sqrt{x-2}}=1

\sqrt{x-2}=2

x-2=4

x=6

y=13

Method B

2(y-5)\frac{dy}{dx}=16

\frac{dy}{dx}=\frac{8}{y-5}

\frac{dy}{dx}=1

\frac{8}{y-5}=1

y=13

64=16x-32

x=6

Equation of tangent

y-13=1(x-6)

y=x+7

Axis intercepts:

(0,7)

and

(-7,0)

Distance RS

=\sqrt{7^2+7^2}

=\sqrt{49\times 2}

=7\sqrt{2}

Grading Criteria

T1
Finds equation of tangent and both axis intercepts
with correct $\dfrac{dy}{dx}$ .

Video Explanation

2022 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 42:39)

2022 TWO(e)