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2009
P2 Q1
Topical
Tangents
09 P2 Q1
2009 H2 Mathematics Paper 2 Question 1
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
(ii)
y
=
2
x
−
12
{y = 2x-12}
y
=
2
x
−
12
(iii)
Q
(
−
3
,
−
18
)
{Q \left( - 3, - 18 \right)}
Q
(
−
3
,
−
18
)
Full solutions
(i)
(ii)
d
x
d
t
=
2
t
+
4
d
y
d
t
=
3
t
2
+
2
t
\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 2 t + 4 \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 3 t^2 + 2 t \\ \end{align*}
d
t
d
x
d
t
d
y
=
2
t
+
4
=
3
t
2
+
2
t
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
3
t
2
+
2
t
2
t
+
4
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=\frac{ 3 t^2 + 2 t }{ 2 t + 4 } \\ \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
2
t
+
4
3
t
2
+
2
t
At point
P
,
{P, }
P
,
t
=
2
{t=2}
t
=
2
x
=
12
y
=
12
d
y
d
x
=
2
\begin{align*} x &= 12 \\ y &= 12 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 2 \end{align*}
x
y
d
x
d
y
=
12
=
12
=
2
Tangent at
P
:
{P:}
P
:
y
−
12
=
2
(
x
−
12
)
y
=
2
x
−
24
+
12
\begin{gather*} y - 12 = 2 \left( x - 12 \right) \\ y = 2x-24+12 \\ \end{gather*}
y
−
12
=
2
(
x
−
12
)
y
=
2
x
−
24
+
12
Hence equation of
l
{l}
l
is
l
:
y
=
2
x
−
12
■
l: \; y = 2x-12 \; \blacksquare
l
:
y
=
2
x
−
12
■
(iii)
x
=
t
2
+
4
t
y
=
t
3
+
t
2
l
:
y
=
2
x
−
12
\begin{align} && \quad x &= t^2 + 4 t \\ && \quad y &= t^3 + t^2 \\ && \quad l: \; y &= 2x - 12 \\ \end{align}
x
y
l
:
y
=
t
2
+
4
t
=
t
3
+
t
2
=
2
x
−
12
Substituting
(
1
)
{(1)}
(
1
)
and
(
2
)
{(2)}
(
2
)
into
(
3
)
,
{(3),}
(
3
)
,
t
3
+
t
2
=
2
(
t
2
+
4
t
)
−
12
t
3
−
t
2
−
8
t
+
12
=
0
\begin{gather*} t^3 + t^2 = 2(t^2 + 4 t) - 12 \\ t^3 - t^2 - 8t + 12 = 0 \end{gather*}
t
3
+
t
2
=
2
(
t
2
+
4
t
)
−
12
t
3
−
t
2
−
8
t
+
12
=
0
We note that
t
=
2
{t=2}
t
=
2
is a solution of this equation
t
3
−
t
2
−
8
t
+
12
=
(
t
−
2
)
(
a
t
2
+
b
t
+
c
)
t^3 - t^2 - 8 t + 12 = (t-2)(at^2+bt+c)
t
3
−
t
2
−
8
t
+
12
=
(
t
−
2
)
(
a
t
2
+
b
t
+
c
)
Comparing coefficients,
x
3
:
a
=
1
x
2
:
b
−
2
a
=
−
1
b
=
1
x
0
:
c
=
−
6
\begin{align*} &x^3:& \quad a &= 1 \\ &x^2:& \quad b-2a &= -1 \\ && b &= 1 \\ &x^0:& \quad c &= -6 \\ \end{align*}
x
3
:
x
2
:
x
0
:
a
b
−
2
a
b
c
=
1
=
−
1
=
1
=
−
6
t
3
−
t
2
−
8
t
+
12
=
0
(
t
−
2
)
(
t
2
+
t
−
6
)
=
0
(
t
−
2
)
(
t
−
2
)
(
t
+
3
)
=
0
\begin{align*} t^3 - t^2 - 8 t + 12 &= 0 \\ (t-2)(t^2 + t - 6) &= 0 \\ (t-2)(t - 2)(t + 3) &= 0 \end{align*}
t
3
−
t
2
−
8
t
+
12
(
t
−
2
)
(
t
2
+
t
−
6
)
(
t
−
2
)
(
t
−
2
)
(
t
+
3
)
=
0
=
0
=
0
t
=
2
(NA)
or
t
=
−
3
\begin{align*} t &= 2 \textrm{ (NA)} \quad \textrm{ or} \\ t &= - 3 \end{align*}
t
t
=
2
(NA)
or
=
−
3
When
t
=
−
3
,
{t=- 3,}
t
=
−
3
,
x
=
−
3
y
=
−
18
\begin{align*} x &= - 3 \\ y &= - 18 \end{align*}
x
y
=
−
3
=
−
18
Hence the coordinates of
Q
{Q}
Q
are
Q
(
−
3
,
−
18
)
■
Q \left( - 3, - 18 \right) \; \blacksquare
Q
(
−
3
,
−
18
)
■
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