2016 ONE(d)

Merit
Question
The tangents to the curve y=14(x−2)2y = \frac{1}{4}(x - 2)^2 at points P and Q are perpendicular.
Loading diagram...
Q is the point (6,4)(6, 4).

What is the xx-coordinate of point P?
*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
y=14(x−2)2y=\frac{1}{4}(x-2)^2
dydx=12(x−2)\dfrac{dy}{dx}=\frac{1}{2}(x-2)

At Q (6,4) dydx=12(6−2)=2Q\ (6,4)\ \dfrac{dy}{dx}=\frac{1}{2}(6-2)=2

∴\therefore At P dydx=−12P\ \dfrac{dy}{dx}=-\frac{1}{2}

−12=12(x−2)-\frac{1}{2}=\frac{1}{2}(x-2)
−1=x−2-1=x-2
x=1x=1
Grading Criteria

Achievement (u)

  • At P dydx=−12P\ \dfrac{dy}{dx}=-\frac{1}{2}

Merit (r)

  • Correct solution with correct derivative.

Excellence T1

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Excellence T2

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Exam Paper (2016)Assessment Schedule (2016)
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