Name:
Date:
APC: Second Semester Final Examination
(Review Day 3 of 3)
3.1
Material/Chapters Covered
Chapter 10: Conic Sections
• parabola, ellipse, and hyperbola
–
–
–
–
–
–
–
know
know
know
know
know
know
know
how to graph if given equation (and the reverse)
the standard form for each
how to find the equation if given a description of the conic (i.e., foci, vertices, etc.)
the characteristics of the conics (i.e., axes, foci, vertices, etc.)
how to rigidly translate a conic
how to rotate the axis of a conic
how to write in polar form
• parametric equations
– know how to use a parametric set of equations
– know how to graph a parametric set of equations
– know how to eliminate the parameter
Chapter 11: Systems of Equations
• matrices for linear systems
–
–
–
–
know
know
know
know
how
how
how
how
to
to
to
to
solve using
solve using
solve using
distinguish
an augmented matrix
inverses
Cramer’s rule
between consistent and inconsistent solutions
• matrix algebra
– know how to find the determinant
– know how to find the inverse
– know how to determine if a matrix is singular
• partial fractions
– know how to express any rational function in terms of partial fractions
– know irreducible quadratics
• non-linear systems
– know how to solve using substitution/elimination
– know how to solve graphically
– know how to check solutions
3.2
Review Questions: Chapter 10 and Chapter 11
Unless otherwise stated in a problem, all results should be exact, simplified, rationalized with angles in radians
and should be done without a calculator.
1. Find the vertex, focus, and the equation of the directrix of
3x2 − 4y = 0
5y 2 + 16x = 32
{Answer:
V : (0, 0)
F
1
0,
3
D:y=−
1
3
and
V : (2, 0)
F
6
,0
5
D:x=
14
5
}
2. Solve each of the following system using any matrix method.
2x − 4y
3x − 6y
=
=
6
9
3x + 5y
4x − 2y
3x − 9y
2x − 6y
= 7
= −3
= 6
= 1
{Answer: In each case we create AX = B from the system of equations. For example, for the first system we have
2x − 4y
3x − 6y
=
=
6
9
linear system
GGGGGGGGGGGGGGGGGGGGGGGGGGA
becomes AX = B
A=
2
3
−4
−6
X=
x
y
B=
6
9
• We get a consistent but infinite solution. The determinant of A is zero and the reduced row-echelon form of the
augmented matrix [A|B] is
1 −2 3
0
0 0
This means the solution is written parametrically
y=t
x = 3 + 2t
where t is any real number.
• We get a consistent and unique solution with x = −1/26 and y = 37/26. The determinant of A is −26 and the reduced
row-echelon form of the augmented matrix [A|B] is


1
 1 0 − 26 



37 
0 1
26
• We get an inconsistent solution (no solution). The determinant of A is zero and the reduced row-echelon form of the
augmented matrix [A|B] is
1 −3 0
0
0 1
which says 0 = 1 which makes no sense which is why there is no solution.
}
3. Find the vertices and foci of the ellipse
3x2 + 4y 2 = 12
9(x + 1)2 + 4y 2 = 36
{Answer:
V (2, 0)
V (−2, 0)
F (1, 0)
F (−1, 0)
V (−1, 3)
}
4. Where is the center of 9x2 + 4y 2 − 72x − 24y + 144 = 0?
{Answer: (4, 3)
Note that the conic turns out to be an ellipse which looks like
(x − 4)2
(y − 3)2
+
=1
4
6
}
V (−1, −3)
F (−1,
√
5)
√
F (−1, − 5)
5. Find the vertices and foci of the hyperbola
3x2 − 4y 2 = 12
9(x + 1)2 − 4y 2 = 36
{Answer:
V (2, 0)
V (−2, 0)
√
F ( 7, 0)
√
F (− 7, 0)
V (1, 0)
V (−3, 0)
F (−1 +
√
13, 0)
F (−1 −
√
13, 0)
}
6. PROBLEM REMOVED (topic not covered)
7. Find the equation in standard form for the parabola
(a) focus (0,5), directrix y = −5 (also give location of vertex)
(b) vertex (4,3), directrix x = 6 (also give location of focus)
(c) y 2 + 4x = 2(y + 6) (give location of vertex and focus)
{Answer:
x2 = 20y,
(y − 3)2 = −8(x − 4),
V (0, 0)
F (2, 3)
13
(y − 1)2 = −4 x −
,
4
V
13
,1 ,
4
F
9
,1
4
}
8. Headlights for automobiles have a parabolic surface of x2 = 12y where x and y are measured in centimeters.
Where should the light bulb be placed?
{Answer: three centimeters from bottom of headlight (which is the bottom of the paraboloid)}
9. Find the equation in standard form of the ellipse
(a) foci: (0, 3) and (0, −3); major axis of length 10
(b) major axis endpoints (1, −4) and (1, 8), minor axis of length 8
(c) 4x2 + y 2 + 16y + 124 = 32x
{Answer:
y2
x2
+
=1
25
16
(y − 2)2
(x − 1)2
+
=1
36
16
(y + 8)2
(x − 4)2
+
=1
4
1
}
10. Solve for matrix X
2 1
0
X=
−2 3
1
2
2
3
−1
{Answer: Pre-Multiply both sides by the inverse of the 2 by 2 matrix that is pre-multiplying X above. This gives X all alone
on the left-hand side and gives this on the right-hand side
−1/8 1/2 5/4
1/4
1 1/2
}
11. Find the equation in standard form of the hyperbola
(a) foci (5, 0) and (−5, 0); transverse axis length is 3
(b) center (0, 0); c = 6; e = 2; horizontal focal axis
(c) center (0, 0); c = 6; e = 2; vertical focal axis
(d) 25y 2 − 9x2 − 50y − 54x − 281 = 0
{Answer:
x2
y2
−
=1
9/4
91/4
}
x2
y2
−
=1
9
27
y2
x2
−
=1
9
27
(y − 1)2
(x + 3)2
−
=1
9
25
12. Give the equations in slope-intercept form of the asymptotes of this conic.
(x + 1)2
(y − 2)2
−
=1
4
3
{Answer:
√
Asymptote A:
(y − 2)
=
Asymptote B:
(y − 2)
=
√
3
(x + 1)
2
√
3
−
(x + 1)
2
or
y
=
or
y
=
√
3
4+ 3
x+
2
2
√
√
3
4− 3
−
x+
2
2
}
13. Find the partial fraction decomposition of the rational function
3x − 2
x2 − 3x − 4
{Answer:
3x − 2
2
1
=
+
x2 − 3x − 4
x−4
x+1
}
14. Identify the type of conic
x2 − 4xy + 3x + 25y − 6 = 0
9x2 − 6xy + y 2 − 7x + 5y = 0
{Answer: hyperbola and parabola}
15. What are the coordinates of the vertices and foci of this conic?
9x2 + 16y 2 − 18x + 64y − 71 = 0
{Answer:
V (5, −2)
V (−3, −2)
F (1 +
√
7, −2)
F (1 −
√
7, −2)
}
16. Find scalar x
2 −3 0 0 x = 127 + 44x
2 1
0
x 1 {Answer: −11/2
Note the above reduces to
4x2 + 44x + 121 = 0
which becomes
(2x + 11)2 = 0
so x = −11/2. }
17. Around the year 1605, German astronomer Johannes Kepler (1571–1630) discovered experimentally what is
now called Kepler’s first law of planetary motion. This law (later proven by Newton using physics) states
that all orbits of planets are elliptical in shape with the sun at a focus of the ellipse. It is even possible to
have binary solar systems with two suns: one at each focus of the ellipse. However, Kepler’s law is even more
general: the thing doing the orbiting can be anything including moons, comets, or satellites and the thing
being orbited can be anything including planets, moons, or suns. Kepler has an asteroid named after him as
well as a crater on both the Moon and on Mars. On the other hand, the location of his grave was lost.
The Moon’s furthest distance from the Earth is 0.2527 million miles and its closest distance from
the Earth is 0.2215 million miles. Use these to find a, b, c, and e for the Moon’s orbit. Comment
on the value of e. Calculator allowed on this problem.
{Answer: In millions of miles the values of a, b, and c are a = 0.2371, b = 0.236586 ≈ 0.2366, c = 0.0156. The eccentricity of the
Moon’s orbit is only e = 0.065795 ≈ 0.066 which means the orbit is nearly circular. This is not unusual. In fact, nearly all the
planets in our solar system have nearly circular orbits about our sun. Only the planet Mercury and the “planet” Pluto have
orbits that really look elliptical (e > 0.2). Halley’s Comet has very, very non-circular orbit about our sun.}
18. Find the polar equation of the conic with a focus at the pole and e = 7/3 and directrix y = −1.
{Answer:
r=
7
3 − 7 sin(θ)
}
19. Find scalars {w, x, y, z} such that A2 = 2AB
x y
1 2
A=
B=
−4 6
w z
{Answer: The answers are {−2, 2, 4, 3}
The long way to do this problem is to find A2 , then find 2AB, then set them equal. This will give four equations for the four
constants.
The short way to do this problem is to begin by manipulating the original equation
A2 = 2AB
−→
A2 − 2AB = 0
−→
A(A − 2B) = 0
In the last step, we note that A is premultiplying in both terms so we can factor it out. The equation
A(A − 2B) = 0
States that either A = 0 (which it can’t since its bottom row is not zero) or A = 2B. This generates the equations for the four
constants very quickly. }
20. Identify the type of conic
√
2x2 + 3 xy + y 2 − 10 = 0
{Answer: A = 2, B =
√
3, C = 1, D = E = 0, and F = −10 ellipse because B 2 − 4AC < 0}