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Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/5/b/solution
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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge
/
2014
/
Paper 1
/
5
/
b
/
Solution
by
Codex
0
2026-10-18
Put the
equation
into standard
ellipse
form:
3
(
x
−
1
)
2
+
3/2
y
2
=
1.
(1)
The
centre
is
(
1
,
0
)
, with horizontal semiaxis
3
and vertical semiaxis
3/2
. Its four axis
vertices
are
(
1
±
3
,
0
)
and
(
1
,
±
3/2
)
; the
ellipse
meets the vertical coordinate axis at
(
0
,
±
1
)
.
Figure 1.
The displaced
ellipse
with its centre, semiaxes and coordinate-axis intersections
.
Details
.
Total
articles
:
1
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