Math Repository
about
topic
al
year
ly
Yearly
2014
P2 Q1
Topical
Tangents
14 P2 Q1
2014 H2 Mathematics Paper 2 Question 1
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
t
=
2.5
{t=2.5}
t
=
2.5
(ii)
x
=
2
27
y
2
{x=\frac{2}{27}y^2}
x
=
27
2
y
2
Full solutions
(i)
d
x
d
t
=
6
t
d
y
d
t
=
6
\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 6 t \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 6 \\ \end{align*}
d
t
d
x
d
t
d
y
=
6
t
=
6
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
6
6
t
=
1
t
\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{6}{6 t} \\ &= \frac{1}{t} \\ \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
6
t
6
=
t
1
When the tangent to
C
{C}
C
has gradient
0.4
{0.4}
0.4
1
t
=
0.4
t
=
2.5
■
\begin{align*} \frac{1}{t} &= 0.4 \\ t &= 2.5 \; \blacksquare \end{align*}
t
1
t
=
0.4
=
2.5
■
(ii)
Tangent at
P
(
3
p
2
,
6
p
)
:
{P(3p^2, 6p): }
P
(
3
p
2
,
6
p
)
:
y
−
6
p
=
1
p
(
x
−
3
p
2
)
y
=
1
p
x
−
3
p
+
6
p
y
=
1
p
x
+
3
p
\begin{gather*} y - 6p = \frac{1}{p} (x- 3p^2) \\ y = \frac{1}{p}x - 3p + 6p \\ y = \frac{1}{p}x + 3p \end{gather*}
y
−
6
p
=
p
1
(
x
−
3
p
2
)
y
=
p
1
x
−
3
p
+
6
p
y
=
p
1
x
+
3
p
When the tangent meets the
y
-
{y\textrm{-}}
y
-
axis,
x
=
0
,
{x=0,}
x
=
0
,
y
=
3
p
D
(
0
,
3
p
)
\begin{gather*} y = 3p \\ D(0, 3p) \end{gather*}
y
=
3
p
D
(
0
,
3
p
)
Mid-point of
P
D
:
{PD: }
P
D
:
(
3
p
2
2
,
9
p
2
)
\left( \frac{3p^2}{2}, \frac{9p}{2} \right)
(
2
3
p
2
,
2
9
p
)
x
=
3
p
2
2
y
=
9
p
2
p
=
2
y
9
\begin{align} && \quad x &= \frac{3p^2}{2} \\ && \quad y &= \frac{9p}{2} \\ && \quad p &= \frac{2y}{9} \\ \end{align}
x
y
p
=
2
3
p
2
=
2
9
p
=
9
2
y
Substituting
(
3
)
{(3)}
(
3
)
into
(
1
)
,
{(1),}
(
1
)
,
x
=
3
(
2
y
9
)
2
÷
2
=
2
27
y
2
\begin{align*} x &= 3 \left( \frac{2y}{9} \right)^2 \div 2 \\ &= \frac{2}{27}y^2 \end{align*}
x
=
3
(
9
2
y
)
2
÷
2
=
27
2
y
2
Cartesian equation of the locus of the mid-point of
P
D
{PD}
P
D
as
p
{p}
p
varies:
x
=
2
27
y
2
■
x=\frac{2}{27}y^2 \; \blacksquare
x
=
27
2
y
2
■
Back to top ▲