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2017
P2 Q1
Topical
Tangents
17 P2 Q1
2017 H2 Mathematics Paper 2 Question 1
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
2
15
units
{2 \sqrt{15} \textrm{ units}}
2
15
units
(ii)
x
y
=
6
{xy=6}
x
y
=
6
Full solutions
(i)
x
=
3
t
y
=
2
t
y
=
2
x
\begin{align} && \quad x = \frac{3}{t} \\ && \quad y = 2 t \\ && \quad y=2x \end{align}
x
=
t
3
y
=
2
t
y
=
2
x
Substituting
(
1
)
{(1)}
(
1
)
and
(
2
)
{(2)}
(
2
)
into
(
3
)
,
{(3),}
(
3
)
,
2
t
=
2
(
3
t
)
2
t
=
6
t
t
2
=
3
t
=
±
3
\begin{align*} 2t &= 2 \left( \frac{3}{t} \right) \\ 2t &= \frac{6}{t} \\ t^2 &= 3 \\ t &= \pm \sqrt{3} \end{align*}
2
t
2
t
t
2
t
=
2
(
t
3
)
=
t
6
=
3
=
±
3
x
=
±
3
3
y
=
±
2
3
\begin{align*} x &= \pm \frac{3}{\sqrt{3}} \\ y &= \pm 2 \sqrt{3} \end{align*}
x
y
=
±
3
3
=
±
2
3
A
(
3
3
,
2
3
)
B
(
−
3
3
,
−
2
3
)
\begin{align*} & A \left( \frac{3}{\sqrt{3}}, 2\sqrt{3} \right) \\ & B \left( -\frac{3}{\sqrt{3}}, -2\sqrt{3} \right) \\ \end{align*}
A
(
3
3
,
2
3
)
B
(
−
3
3
,
−
2
3
)
Exact length of
A
B
=
(
3
3
+
3
3
)
2
+
(
2
3
+
2
3
)
2
=
36
3
+
16
(
3
)
=
60
=
2
15
units
■
\begin{align*} & \textrm{Exact length of } AB \\ &= \sqrt{ \left( \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}} \right)^2 + \left( 2\sqrt{3} + 2\sqrt{3} \right)^2 } \\ &= \sqrt{ \frac{36}{3} + 16(3) } \\ &= \sqrt{ 60 } \\ &= 2 \sqrt{15} \textrm{ units} \; \blacksquare \end{align*}
Exact length of
A
B
=
(
3
3
+
3
3
)
2
+
(
2
3
+
2
3
)
2
=
3
36
+
16
(
3
)
=
60
=
2
15
units
■
(ii)
d
x
d
t
=
−
3
t
2
d
y
d
t
=
2
\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= - \frac{3}{t^{2}} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 2 \\ \end{align*}
d
t
d
x
d
t
d
y
=
−
t
2
3
=
2
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
2
÷
(
−
3
t
2
)
=
−
2
3
t
2
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=2 \div \left(- \frac{3}{t^{2}}\right) \\ &=- \frac{2}{3} t^2 \\ \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
2
÷
(
−
t
2
3
)
=
−
3
2
t
2
At point
P
(
3
p
,
2
p
)
,
{P \left( \frac{3}{p}, 2p \right), }
P
(
p
3
,
2
p
)
,
t
=
p
{t=p}
t
=
p
d
y
d
x
=
−
2
3
p
2
\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{2}{3} p^2
d
x
d
y
=
−
3
2
p
2
Tangent at
P
(
p
2
,
2
p
)
:
{P\left( p^2, \frac{2}{p} \right):}
P
(
p
2
,
p
2
)
:
y
−
2
p
=
−
2
3
p
2
(
x
−
3
p
)
y
=
−
2
3
p
2
x
+
2
p
+
2
p
y
=
−
2
3
p
2
x
+
4
p
\begin{gather*} y - 2p = - \frac{2}{3} p^2 \left( x - \frac{3}{p} \right) \\ y = - \frac{2}{3} p^2 x + 2p +2p \\ y = - \frac{2}{3} p^2 x + 4p \\ \end{gather*}
y
−
2
p
=
−
3
2
p
2
(
x
−
p
3
)
y
=
−
3
2
p
2
x
+
2
p
+
2
p
y
=
−
3
2
p
2
x
+
4
p
At
D
,
y
=
0
{D, y=0}
D
,
y
=
0
0
=
−
2
3
p
2
x
+
4
p
x
=
6
p
\begin{align*} 0 &= - \frac{2}{3} p^2 x + 4p \\ x &= \frac{6}{p} \end{align*}
0
x
=
−
3
2
p
2
x
+
4
p
=
p
6
At
E
,
x
=
0
{E, x=0}
E
,
x
=
0
y
=
−
2
3
p
2
(
0
)
+
4
p
=
4
p
\begin{align*} y &= - \frac{2}{3} p^2 (0) + 4p \\ &= 4p \end{align*}
y
=
−
3
2
p
2
(
0
)
+
4
p
=
4
p
D
(
6
p
,
0
)
E
(
0
,
4
p
)
F
(
3
p
,
2
p
)
\begin{align*} & D \left( \frac{6}{p}, 0 \right) \\ & E \left( 0, 4p \right) \\ & F \left( \frac{3}{p}, 2p \right) \\ \end{align*}
D
(
p
6
,
0
)
E
(
0
,
4
p
)
F
(
p
3
,
2
p
)
x
=
3
p
y
=
2
p
p
=
y
2
\begin{align} &&\quad x &= \frac{3}{p} \\ &&\quad y &= 2p \\ &&\quad p &= \frac{y}{2} \end{align}
x
y
p
=
p
3
=
2
p
=
2
y
Substituting
(
6
)
{(6)}
(
6
)
into
(
4
)
,
{(4),}
(
4
)
,
x
=
3
÷
y
2
=
6
y
\begin{align*} x &= 3 \div \frac{y}{2} \\ &= \frac{6}{y} \end{align*}
x
=
3
÷
2
y
=
y
6
Cartesian equation of the curve traced by
F
{F}
F
as
p
{p}
p
varies:
x
y
=
6
■
xy=6 \; \blacksquare
x
y
=
6
■
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