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ASSESSMENT SCHEDULE (Sample)
Mathematics with Calculus: Sketch graphs and find equations of conic sections (90639)
Achievement
Criteria
Sketch graphs
of conic
sections.
No
One
Evidence
Code
Judgement
x2 + (y + 5)2 = 9
ACHIEVEMENT:
Centre and
intercepts
indicated by
sketch.
y
– 4– 3– 2– 1
1
Sufficiency
2
3
4
x
– 2
– 4
A1
– 6
Circle drawn
through
(0, –2),
(0, –8).
Two of
Code A1
and
Two of
Code A2
– 8
ACHIEVEMENT
Two
x2 y2
+
=1
16 4
Centre and
intercepts
indicated by
sketch.
y
4
2
– 4
– 2
2
4
x
A1
– 2
– 4
Three
Ellipse
through
(–4,0), (0,2),
(4,0), (0,–2).
(y − 2) 2 = 4x
y
Vertex and
intercepts
indicated by
sketch.
5
4
3
2
1
– 1
1
2
3
4
x
A1
Parabola
through (1,0),
(0,2)
 New Zealand Qualifications Authority, 2003
All rights reserved. This publication may be reproduced for use with students and the training of teachers for the implementation of NCEA.
Achievement
Criteria
ACHIEVEMENT
Find
equations of
conic sections
from given
information.
No
Four
(a)
Evidence
A2
Or equivalent.
A2
Or equivalent.
A2
Or equivalent.
x2 y2
−
=1
36 b2
100 16
−
=1
36 b2
16 16
= 2
9
b
x2 y2
−
=1
36
9
Two of
Code A1
and
Hyperbola:
b2 = 9
∴ Equation of Hyperbola:
Sufficiency
ACHIEVEMENT:
Ellipse:
(x − 3)2 (y − 4)2
+
=1
9
4
(c)
Judgement
Circle:
(x – 4)2 + (y – 2)2 = 9
(b)
Code
Mathematics with Calculus 90639 Assessment Schedule (sample) Page 2 of 5
Two of
Code A2
Achievement
Criteria
Solve
problems
involving
conic
sections.
No
Five
Evidence
Code
x = 2cos 2t
Judgement
y = 2sin t
dx
= − 4 sin 2t
dt
Merit:
Achievement
dy
= 2 cos t
dt
plus
Two of
Code M
dy dy dt
= ×
dx dt dx
1
1
=− ×
4 sin t
At (1,1)
or
sin t =
dy
1
=− .
dx
2
All three
Code M
1
2
dy
dx
Stated.
MERIT
Equation of tangent:
x + 2y − 3 = 0
or
y=−
1
3
x+
2
2
OR
x = 2cos 2t
y = 2sin t
y2 = 2 − x
Differentiating implicitly:
2y ×
dy
= −1
dx
dy −1
=
dx 2y
At (1,1):
Equation:
Gradient of
tangent
evaluated.
dy
1
=− .
dx
2
y−1
1
=−
x−1
2
M
x + 2y − 3 = 0
or
Sufficiency
1
2
y=– x+
Equation of
tangent
written.
3
2
Mathematics with Calculus 90639 Assessment Schedule (sample) Page 3 of 5
Achievement
Criteria
Solve conic
section
problems.
Evidence
No
Six
Code
Equation of circle:
2
x + y = 196
y
Two of
Code M
– 14 – 7
–7
7
14
x
Graph drawn.
Radius of
base
calculated
and
substituted
into area
formula.
Solve for x when y = –12
x 2 + (−12)2 = 196
x 2 = 52
x=
52
2
Area of base = πr
= 52π ≈ 163.4 cm2
MERIT
M
Vertex is (0,2.5) so equation of
parabola is: x 2 = k(y − 2.5)
Or
equivalent.
Units not
required.
y
8
6
4
2
– 20
or
All three
Code M
– 14
– 30
Achievement
plus
7
7
Sufficiency
MERIT:
Equation of
circle stated.
2
14
Judgement
– 10
10
x
A1
Graph drawn.
A2
Equation of
parabola
stated.
Equation of parabola:
x2 = 200 (y – 2.5)
or
y = 0.005x2 + 2.5
y = 3.48 at right hand post:
x2 = 200 (3.48 – 2.5)
x = 14
The posts are 44 metres apart.
M
Distance
between the
posts is
clearly stated.
Mathematics with Calculus 90639 Assessment Schedule (sample) Page 4 of 5
Achievement
Criteria
Evidence
No
Solve more
Eight
difficult conic
(a)
section
problems.
Code
Differentiating implicitly gives:
Merit
Gradient
function is
found.
dy −b cosθ
=
dx
a sin θ
Equation of normal:
y − b sinθ
a sinθ
=
x − a cosθ b cosθ
2
by cosθ − b sinθ cosθ = ax sinθ −
a2sinθ cosθ
M
E
Point and
normal
gradient are
substituted.
ax sinθ − by cosθ = (a2 − b2) sinθ cosθ.
(b)
According to the result from Eight
(a):
Any normal to
x2 y2
+
= 1 has the
25 9
equation 5x sinθ − 3y cosθ = 16sinθ cosθ.
A: x-intercept: (
Sufficiency
EXCELLENCE:
dy −b 2 x
= 2
dx
a y
At P(acosθ, bsinθ):
EXCELLENCE
Judgement
16
cosθ , 0)
5
B: y-intercept: (0,
−16
sin θ)
3
8
−8
5
3
Midpoint of AB: ( cos θ ,
sin θ )
Equation of ellipse formed by the
locus of midpoints of AB:
A2 M
E
Equation of
ellipse stated.
25 x 2 9 y 2
+
= 1 or 25x2 + 9y2 = 64
64
64
Mathematics with Calculus 90639 Assessment Schedule (sample) Page 5 of 5
plus
Both Code E