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2010
P1 Q4
Topical
Tangents
10 P1 Q4
2010 H2 Mathematics Paper 1 Question 4
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
d
y
d
x
=
x
+
y
y
−
x
{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x+y}{y-x}}
d
x
d
y
=
y
−
x
x
+
y
(ii)
(
2
,
−
2
)
,
(
−
2
,
2
)
{\left(\sqrt{2}, -\sqrt{2}\right), \;}\allowbreak {\left(-\sqrt{2}, \sqrt{2}\right)}
(
2
,
−
2
)
,
(
−
2
,
2
)
Full solutions
(i)
x
2
−
y
2
+
2
x
y
+
4
=
0
\begin{equation} \qquad x^2 - y^2 + 2xy + 4 = 0 \end{equation}
x
2
−
y
2
+
2
x
y
+
4
=
0
2
x
−
2
y
d
y
d
x
+
2
x
d
y
d
x
+
2
y
=
0
2x - 2y \frac{\mathrm{d}y}{\mathrm{d}x} + 2x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 0
2
x
−
2
y
d
x
d
y
+
2
x
d
x
d
y
+
2
y
=
0
d
y
d
x
(
2
x
−
2
y
)
=
−
2
x
−
2
y
d
y
d
x
=
x
+
y
y
−
x
■
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \left( 2x-2y \right) &= -2x-2y \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{x+y}{y-x} \; \blacksquare \end{align*}
d
x
d
y
(
2
x
−
2
y
)
d
x
d
y
=
−
2
x
−
2
y
=
y
−
x
x
+
y
■
(ii)
When the tangent is parallel to the
x
-
{x\textrm{-}}
x
-
axis,
d
y
d
x
=
0
x
+
y
y
−
x
=
0
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 0 \\ \frac{x+y}{y-x} &= 0\\ \end{align*}
d
x
d
y
y
−
x
x
+
y
=
0
=
0
y
=
−
x
\begin{equation} \qquad y = -x \end{equation}
y
=
−
x
Substituting
(
2
)
{(2)}
(
2
)
into
(
1
)
,
{(1),}
(
1
)
,
x
2
−
(
−
x
)
2
+
2
x
(
−
x
)
+
4
=
0
2
x
2
=
4
\begin{gather*} x^2 - (-x)^2 + 2x(-x) + 4 = 0 \\ 2x^2 = 4 \end{gather*}
x
2
−
(
−
x
)
2
+
2
x
(
−
x
)
+
4
=
0
2
x
2
=
4
x
=
±
2
y
=
∓
2
\begin{align*} x &= \pm \sqrt{2} \\ y &= \mp \sqrt{2} \end{align*}
x
y
=
±
2
=
∓
2
Coordinates of points at which the tangent is parallel to the
x
-
{x\textrm{-}}
x
-
axis:
(
2
,
−
2
)
■
,
(
−
2
,
2
)
■
\begin{align*} \left(\sqrt{2}, -\sqrt{2}\right) \;& \blacksquare, \\ \left(-\sqrt{2}, \sqrt{2}\right) \; &\blacksquare \end{align*}
(
2
,
−
2
)
(
−
2
,
2
)
■
,
■
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