2017 ONE(c)

Merit
Question
The normal to the parabola y=0.5(x−3)2+2y = 0.5(x - 3)^2 + 2 at the point (1,4)(1,4) intersects the parabola again
at the point P.
Loading diagram...
Find 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=0.5(x−3)2+2y=0.5(x-3)^2+2
dydx=2×0.5×(x−3)\dfrac{dy}{dx}=2\times0.5\times(x-3)
=x−3\qquad =x-3
At x=1x=1 dydx=−2\dfrac{dy}{dx}=-2

∴\therefore For normal dydx=12\dfrac{dy}{dx}=\dfrac{1}{2}
Through (1,4)(1,4) ∴\therefore Eqn of normal y=12x+3.5\qquad y=\dfrac{1}{2}x+3.5

At point P: 12x+3.5=0.5(x−3)2+2\qquad \dfrac{1}{2}x+3.5=0.5(x-3)^2+2

x+7=(x−3)2+4\qquad x+7=(x-3)^2+4
x+7=x2−6x+9+4\qquad x+7=x^2-6x+9+4
x2−7x+6=0\qquad x^2-7x+6=0
(x−6)(x−1)=0\qquad (x-6)(x-1)=0
At point P x=6\qquad x=6
Grading Criteria

Achievement (u)

  • Correct expression for dydx\dfrac{dy}{dx}
    (i.e. correct derivative).

Merit (r)

  • Correct solution with correct derivative.

Excellence T1

-

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2017 NZQA Exam - Worked Answers by infinityplusone(starts at 5:12)
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Exam Paper (2017)Assessment Schedule (2017)
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