Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Centre for Education in
Mathematics and Computing
Grade 12 Math Circles
November 25, 2015
Solutions
1. (a) We require,
x2 + y 2 = (3x + 2)2 or x2 + y 2 = 9x2 + 12x + 4
This gives,
8x2 + 12x − y 2 + 4 = 0
This can be rewritten as,
3
3
3
3
8(x2 + x) − y 2 + 4 = 0 or 8(x2 + x + ( )2 ) − y 2 + 4 = 8( )2
2
2
4
4
Thus we obtain
3
3
9
1
3
8(x + )2 − y 2 = 8( )2 − 4 or 8(x + )2 − y 2 = − 4 =
4
4
4
2
2
Thus, the conic section is the following hyperbola:
(x + 43 )2
y2
−
=1
(1/16)
(1/2)
(b) Similarly, we write
x
x2
x2 + y 2 = ( + 3)2 or x2 + y 2 =
+ 3x + 9
2
4
This gives,
3x2 − 12x + 4y 2 = 36
1
This can be rewritten as,
3(x2 − 4x) + 4y 2 = 36 or 3(x2 − 4x + 4) + 4y 2 = 36 + 12
We obtain
3(x − 2)2 + 4y 2 = 48
Thus, the conic section is the following ellipse:
(x − 2)2 y 2
+
=1
16
12
2. (a) The general equation of the parabola is y 2 = 4px (p > 0). Let F (0, p) be the focus.
The opening at the focus is 10 cm, thus the point A(p, 5) is on the parabola.
Thus, we have
52 = 4p(p)
This gives p2 = 25/4 or p = 5/2. Thus the parabola is given by,
5
y 2 = 4( )x or y 2 = 10x
2
(b) At x = 11, we have
√
y 2 = 10(11) = 110 or y = ± 110
√
Thus, the diameter of the opening at 11 cm is 2 110.
3. (a) We have
x = |P A| − |P B| = vt = (908)(1200) = 117600 (ft)
Thus the distance difference is 1176000 (ft) or 2450/11 (mi). From the definition of a
hyperbola,
|P A| − |P B| = 2a =
2
2450
1225
or a =
11
11
We know that c = 200 (mi), thus
b 2 = c 2 − a2 =
3339375
121
The equation of the hyperbola is
y2
x2
−
=1
(1500625/121) 3339375/121
(b) We have x = 200 (mi), thus
2002
y2
−
= 1 or y ≈ 248 (mi)
1500625/121 3339375/121
4. We saw that for the projectile motion,
y = (tan θ)x −
2v02
g
x2
cos2 θ
If the ball is thrown horizontally, θ = 0, thus we obtain the following parabola:
y=−
g 2
x
2v02
(b) If h = 5 m and v0 = 2 m/s, we have
−5 = −
Thus, we obtain x = 2 m.
3
10 2
x
2(2)2