Q1
r=
p
√
x2 + y 2 , φ = arctan(y/x) ⇒ (r, φ) = ( 2, π/4)
√
x = r cos(φ), y = r sin(φ) ⇒ (x, y) = (−1, 3)
Q2
Figure 1: r =| cos(2θ) |
Figure 2: r =| sin(3θ) |
Q3
x2
(y − 1)2
+
=1
148/9
37/4
√
which is the formula√for an ellipse centered at (0, 1) with major axis of length a = 2 37/3
and minor axis b = 37/2 so:
r
p
√
592 − 333 √
c = a2 − b2 = 148/9 − 37/4 =
= 259/6
36
√
√
So the focii are at (− 259/6, 1) and ( 259/6, 1).
132 = 9x2 +16y 2 −32y = 9x2 +16(y−1)2 −16 ⇔ 9x2 +16(y−1)2 = 148 ⇔
1
Figure 3: 9x2 + 16y 2 − 32y = 132
Q4 This is a hyperbola with a = 3/2 and c = 5 which means that b =
y2
x2
so the equation is 9/4
− 91/4
=1
√
√
c2 − a2 = 91/2
Q5 The origin of the new coordinates is at (2, 1) so there is a shift by (−2, −1) for the new
coordinates. Then the axes are rotated by π/3 counterclockwise, so the new coordinates of
a point are:
X = (x − 2) cos(π/3) + (y − 1) sin(π/3)
Y = −(x − 2) sin(π/3) + (y − 1) cos(π/3)
so:
√
√
(3, 3) → ((x−2) cos(π/3)+(y−1) sin(π/3), −(x−2) sin(π/3)+(y−1) cos(π/3)) = (1/2+ 3, 1− 3/2)
Q6
Figure 4: xy = 1
Rotate coordinates and sub into xy = 1:
√
√
X = x cos(π/4) + y sin(π/4) = (y + x)/ √
2
x = (X − Y )/√ 2
⇒
Y = −x sin(π/4) + y cos(π/4) = (y − x)/ 2
y = (X + Y )/ 2
1 = xy = X 2 /2 − Y 2 /2
Q7 Curve 1: δ = (9)(16) − (−24)2 /4 = 0 so it is a parabola.
Curve 2: δ = (5)(5) − (−6)2 /4 = 16 > 0 so it is an ellipse.
Curve 3: δ = (2)(−1) − (−4)2 /4 = −6 < 0 so it is a hyperbola.
2
cot(2θ) = a−c
= −3/4 ⇒ cos2 (2θ) = 9 sin2 (2θ)/16 = 9/16 − 9 cos2 (2θ)/16 ⇒ cos2 (2θ) =
b
9/25 ⇒ cos(2θ) = ±3/5
We can assume that 0 < θ < π/2 because if θ is wrong by some multiple of π/2 it can
easily be fixed by subbing in (X, Y ) = (±y, ∓x) or (X, Y ) = (−x, −y). Then sin(2θ) > 0
so cot(2θ) has the same sign as cos(2θ) since cot(2θ) = cos(2θ)/ sin(2θ). So we get that
cos(2θ) = −3/5.
√
−3/5 = cos(2θ) = 2 cos2 (θ) − 1 ⇒ cos2 (θ) = 1/5 ⇒ cos(θ)
√ = ±1/ 5
√
Using the fact that 0 < θ < π/2 we get that cos(θ) = 1/ 5 and sin(θ) = 2/ 5 so:
√
√
X = x cos(θ) + y sin(θ) = (x + 2y)/ √
5
x = (X − 2Y )/√ 5
⇒
Y = −x sin(θ) + y cos(θ) = (y − 2x)/ 5
y = (2X + Y )/ 5
Subbing this into the equation gives:
0 =
=
=
⇒1 =
2x2 − 4xy − y 2 + 6
2(X 2 − 4XY + 4Y 2 )/5 − 4(2X 2 − 3XY − 2Y 2 )/5 − (4X 2 + 4XY + Y 2 )/5 + 6
−2X 2 + 3Y 2 + 6
X 2 /3 − Y 2 /2
Q8*(a)
|P F1 | + |P F2 | = r
p
p
(x − f )2 + y 2 +
(x + f )2 + y 2 = r
p
p
(x − f )2 + y 2 + (x + f )2 + y 2 + 2 (x − f )2 + y 2 (x + f )2 + y 2 = r2
p
2 (x2 − f 2 )2 + y 4 + y 2 ((x − f )2 + (x + f )2 )
4((x2 − f 2 )2 + y 4 + y 2 ((x − f )2 + (x + f )2 ))
−8x2 f 2 + 4f 4 + 8y 2 f 2
(4(r2 − 2f 2 ) − 8f 2 )x2 − (4(r2 − 2f 2 ) + 8f 2 )y 2
4(r2 − 4f 2 )x2
r2 (r2 − 4f 2 )
4x2
r2
r2 − 2(x2 + y 2 + f 2 )
(r2 − 2(x2 + y 2 + f 2 ))2
(r2 − 2f 2 )2 − 2(r2 − 2f 2 )(2x2 + 2y 2 )
(r2 − 2f 2 )2 − 4f 2
4r2 y 2
+ 2 2
=1
r (r − 4f 2 )
4y 2
+
=1
(r2 − 4f 2 )
p
whic is the standard form of an ellipse with a = r/2 and b = r2 − 4f 2 /2.
=
=
=
=
Q8*(b)
||P F1 | − |P F2 || = d
p
p
| (x − f )2 + y 2 −
(x + f )2 + y 2 | = d
p
p
(x − f )2 + y 2 + (x + f )2 + y 2 − 2 (x − f )2 + y 2 (x + f )2 + y 2 = d2
p
−2 (x2 − f 2 )2 + y 4 + y 2 ((x − f )2 + (x + f )2 )
4((x2 − f 2 )2 + y 4 + y 2 ((x − f )2 + (x + f )2 ))
−8x2 f 2 + 4f 4 + 8y 2 f 2
(4(d2 − 2f 2 ) − 8f 2 )x2 − (4(d2 − 2f 2 ) + 8f 2 )y 2
4(d2 − 4f 2 )x2
d2 (d2 − 4f 2 )
4x2
d2
4x2
d2
3
d2 − 2(x2 + y 2 + f 2 )
(d2 − 2(x2 + y 2 + f 2 ))2
(d2 − 2f 2 )2 − 2(d2 − 2f 2 )(2x2 + 2y 2 )
(d2 − 2f 2 )2 − 4f 2
4d2 y 2
+ 2 2
=1
d (d − 4f 2 )
4y 2
+
=1
(d2 − 4f 2 )
4y 2
−
= 1 (since d < 2f )
(4f 2 − d2 )
=
=
=
=
which is the standard form of a hyperbola with a = d/2 and b =
p
4f 2 − d2 /2.
Q8*(c)
|P F |
x + (y − f )2
x2
x2
2
|y + f |
(y + f )2
y 2 + 2yf + f 2 − (y 2 − 2yf + f 2 ) = 0
4f y
x∗2
y =
4f
=
=
=
=
which is the standard form of a parabola with a = 1/4f .
4