MEI Core 4

Parametric Equations

Chapter Assessment

1. A curve is defined by the parametric equations x = 2t 2 , y = 4t . (i) By eliminating the parameter, find the cartesian equation of the curve. (ii) Find the equation of the tangent to the curve at the point A with parameter t = 2. (iii) Show that the tangent does not meets the curve again. (iv) The normal of the curve at A cuts the curve again at B. Find the coordinates of B. [3] [4] [3] [5]

2. Find the turning points of the curve with parametric equations x = 3t , y = 12t − t 3 and distinguish between them. [5] 3. A circle is defined by the parametric equations x = 1 + 2 cos θ , y = 3 + 2sin θ . (i) Sketch the circle. dy (ii) Find at the point with parameter θ . dx (iii) Find the equation of the tangent at the point with parameter θ . π (iv) Find the coordinates of the point where θ = . 3 π (v) Find the equation of the normal at the point where θ = . 3

[2] [3] [3] [2] [4]

4. A line is defined by the parametric equations x = cos 2t , y = sin 2 t dy . (i) Find dx (ii) Find the cartesian equation of the line.

[3] [3]

Total 40 marks

© MEI, 01/09/08

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MEI C4 Parametric Equations Assessment solutions

Solutions to Chapter Assessment

1. (i)

x = 2t 2

y = 4t ⇒ t =

Insert into

y

4

:

2 y2 ⎛y⎞ x = 2⎜ ⎟ ⇒ x = 8 ⎝4⎠ 2 ⇒ y = 8x

(ii) x = 2t 2 ⇒

dx = 4t dt dy y = 4t ⇒ =4 dt dy dy / dt 4 1 = = = dx dx / dt 4t t dy 1 = When t = 2: dx 2

[3]

When t = 2: x = 2 × 2 2 = 8 and y = 4 × 2 = 8

1 So the tangent has gradient 2 and passes through (8, 8) 1 Equation of tangent is y − 8 = 2 ( x − 8 ) 2 y − 16 = x − 8 2y = x + 8

(iii) Where tangent meets curve, 2 × 4t = 2t + 8

2

[4]

2t 2 − 8t + 8 = 0

t 2 − 4t + 4 = 0

(t − 2)2 = 0 The only root is the repeated root at t = 2 (the point of the tangent) so the tangent doesn’t meet the curve again. [3] 1 (iv) Gradient of tangent = 2 , so gradient of normal = −2 (using m 1m2 = −1 ) Normal has gradient -2 and passes through (8, 8)...