MATHEMATICS EXTENSION 2
EXERCISE 32_123
/
4 UNIT MATHEMATICS
TOPIC 3: CONICS
(HSC 2013/2)
Consider the equation
4 x 2  25 y 2  100
Part (A) What type of curve does the equation correspond too? Is the eccentricity e of the
curve: e = 1; e < 1, e > 1; e = 0 ?
Part (B) Give the Cartesian coordinates for the vertices and focal points. Calculate the
eccentricity e of the curve?
Part (C)
State the equations for the directrices and asymptotes of the curve.
Part (D) The point P on the curve has the X-coordinate xP = 10 and yP > 0. Where does the tangent to the curve
at the point P cut the X-axis and Y-axis? Where does the normal to the tangent at the point P intersect
the X-axis and the Y-axis
Part (E)
Sketch the curve showing the vertices, focal points, the asymptotes, directrices, and the points where
the tangent and the normal intersect the X-axis and Y-axis.
For your sketch: X-axis (-30 to +30) and Y-axis (-30 to +30).
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Answer Part (A)
The equation 4 x 2  25 y 2  100 corresponds to the curve of a hyperbola.
The eccentricity of hyperbolas is e > 1.
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Answer Part (B)
x2 y 2
A general expression for a hyperbola is 2  2  1
a
b
The equation of the hyperbola
4 x 2  25 y 2  100 can be re-written as
x2 y 2

1  a  5 b  2
52 22
The vertices of the hyperbola are A1(-a, 0) and A2(a, 0)
A1(-5,0) and A2(5,0)
The focal length c is c 2  a 2  b2
c  a 2  b2  25  4  29  5.3852
The focal points are F1(-c, 0) and F2(c, 0)
F1(- 29 ,0) and F2( 29 ,0)
The eccentricity is e 
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or
F1(-5.3852, 0) and F2(5.3852, 0)
c
29

 1.0770  1
a
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Answer Part (C)
The equations for the directrices are
x
a2
c

x
25
 4.6424
29
x
25
 4.6424
29
The equations for the asymptotes are
y
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b
x 
a
y
2
x  0.4000 x
5
y
2
x  0.4000 x
5
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Answer Part (D)
The coordinates of the point P are (xP, yP)
xP  10
 100 
yP  4 
  1  12  2 3
 25 
The equation of the straight line for the tangent is
y  M 1 x  B1
The gradient of the curve is given by the first derivative of the function
 2 
 2   dy 
 x   y    0
 25 
 4   dx 
 4  x 
dy / dx    
 25   y 
The gradient M1 at the point P
xP  10
 4   10   4 
yP  2 3 M 1    

  0.4619
 25   2 3   5 3 
The intercept B1 of the tangent is
 4  2
B1  yP  M1 xP  2 3  10 
 1.1547

3
5 3
The tangent intersects the X-axis at the point T
yT  0 xT  
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B1
 2   5 3 
 
  2.5

M1
 3  4 
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The tangent intersects the Y-axis at the Point U
xU  0
yU  B1 
2
 1.1547
3
The equation of the straight line for the normal is
where
M2 
y  M 2 x  B2
5 3
1
 
  2.1651
M1
 4 
The intercept B2 of the normal is
5 3
B2  yP  M 2 xP  2 3  10 
  14.5 3  25.1147
4


The normal intersects the X-axis at the point R
yR  0 xR  
B2
 4 
  14.5 3 
  11.600
M2
5 3


The normal intersects the Y-axis at the Point S
xS  0
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yS  B2  14.5 3  25.1147
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Answer Part (E)
In the graphs, the vertices and focal points (blue dots) are very close to each other
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