MA 22S1
Assignment 3
Due 2-4 November 2015
Id:
22S1-f2015-3.m4,v 1.1 2015/10/26 15:30:43 john Exp john
1. For each of the following, determine whether it is an ellipse, parabola,
hyperbola or degenerate. In the degenerate case you don’t need to
specify which kind of degenerate conic it is.
(a) x2 + 2xy + 3y 2 + 4x + 5y + 6 = 0
Solution: This is in the form Ax2 + Bxy + Cy 2 + Dx+ Ey + F = 0.
In other words it is a conic, so we can classify it using the invariants
τ = A + C,
δ = B 2 − 4AC,
ǫ = 8ACF + 2BED − 2AE 2 − 2CD 2 − 2B 2 F.
In this case
τ = 4,
δ = −8,
ǫ = 30.
δ < 0, but τ ǫ ≥ 0, so this is degenerate.
(b) x2 + 2xy + 3y 2 + 4x + 5y − 6 = 0
Solution:
τ = 4, δ = −8,
ǫ = −162.
δ < 0 and τ ǫ < 0, so this is an ellipse.
(c) 3x2 − 4xy − 4y 2 + 10x + 4y − 3 = 0
Solution:
τ = −1, δ = 64,
ǫ = 768.
δ > 0 and ǫ 6= 0, so this is a hyperbola.
1
Id:
22S1-f2015-3.m4,v 1.1 2015/10/26 15:30:43 john Exp john 2
(d) 3x2 − 4xy − 4y 2 + 10x + 4y + 3 = 0
Solution:
τ = −1, δ = 64,
ǫ = 0.
δ < 0, but ǫ = 0, so this is degenerate.
(e) x2 − 4xy + 4y 2 + 6x − 12y + 4 = 0
Solution:
τ = 5, δ = 0,
ǫ = 0.
δ = 0, but τ ǫ ≥ 0, so this is degenerate.
(f) x2 − 4xy + 4y 2 + 3x − 14y + 8 = 0
Solution:
τ = 5, δ = 0, ǫ = −128.
δ = 0 and τ ǫ < 0, so this is a parabola.
2. The set of (x, y) where
94x2 − 360xy − 263y 2 + 572x + 494y − 338 = 0
is a hyperbola.
(a) Put it in standard form via an appropriate change of variables.
Solution: First we rotate through an angle of
ϕ=
B
1
−360
1
arctan
= arctan
= −0.39479
2
A−C
2
94 − −263
cos ϕ = 0.92308 = 12/13 and sin ϕ = −0.38462 = −5/13, so the
first change of variable we want is
x′ =
x=
12
5
x − y,
13
13
12 ′
5
x + y′,
13
13
y′ =
5
12
x + y,
13
13
y=−
5 ′ 12 ′
x + y.
13
13
Substituting,
94x2 − 360xy − 263y 2 + 572x + 494y − 338
= 169(x′ )2 − 338(y ′)2 + 338x′ + 676y ′ − 338.
We then translate by (h, k), where
h=
1 338
= 1,
2 169
k=
1 676
= −1.
2 −338
Id:
22S1-f2015-3.m4,v 1.1 2015/10/26 15:30:43 john Exp john 3
The second change of variable is
Substituting again,
x′′ = x′ + 1,
y ′′ = y ′ − 1,
x′ = x′′ − 1,
y ′ = y ′′ + 1.
169(x′ )2 − 338(y ′)2 + 338x′ + 676y ′ − 338.
= 169(x′′ )2 − 338(y ′′)2 − 169.
So our curve is
169(x′′ )2 − 338(y ′′)2 − 169 = 0
or, more simply,
(x′′ )2 − 2(y ′′)2 = 1.
This is indeed a hyperbola, as promised. It is
(x′′ )2 (y ′′)2
− 2 =1
a2
b
√
where a = 1 and b = 1/ 2.
(b) Find its eccentricity.
Solution:
√
b2
=
e = c/a =
q
c=
a2
+
and
q
3/2
3/2.
(c) Find its asymptotes in the original coordinate system.
Solution: In the new coordinate system the asymptotes are
√
b
y ′′ = ± x′′ = ± 2x′′ .
a
Substituting,
√
y ′ − 1 = ± 2(x′ + 1)
√ 12
5
12
5
x+ y−1 =± 2
x− y+1 ,
13
13
13
13
or
√
√
√
(5 ∓ 12 2)x + (12 ± 5 2)y = 13 ± 13 2.
This is already a sufficient answer, but if you want y in terms of
x then
√
√
(−60 ± 169 2)x + 26 ± 91 2
y=
.
94