Question

PLEASE HELP
8.08, part 2

11. Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

A) y squared over 45 minus x squared over 36 = 1
B) y squared over 81 minus x squared over 36 = 1
C) y squared over 36 minus x squared over 81 = 1
D) y squared over 36 minus x squared over 45 = 1

12. Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x.

A) y squared over 16 minus x squared over 64 = 1
B) y squared over 16 minus x squared over 256 = 1
C) y squared over 256 minus x squared over 16 = 1
D) y squared over 64 minus x squared over 4 = 1

13. Eliminate the parameter.
x = t - 3, y = t2 + 5

A) y = x2 + 6x + 14
B) y = x2 - 14
C) y = x2 - 6x - 14
D) y = x2 + 14

14. Find the rectangular coordinates of the point with the polar coordinates.
ordered pair 3 comma 2 pi divided by 3

A) ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2
B) ordered pair 3 square root 3 divided by 2 comma negative 3 divided by 2
C) ordered pair negative 3 divided by 2 comma 3 divided by 2
D) ordered pair 3 divided by 2 comma negative 3 divided by 2

15. Find all polar coordinates of point P where P = negative pi divided by 6 .

A) (1, negative pi divided by 6 + (2n + 1)π) or (-1, negative pi divided by 6 + 2nπ)
B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ)
C) (1, negative pi divided by 6 + 2nπ) or (1, pi divided by 6 + (2n + 1)π)
D) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Determine two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

A) (4 square root 2 , 135°), (-4 square root 2 , 315°)
B) (4 square root 2 , 45°), (-4 square root 2 , 225°)
C) (4 square root 2 , 315°), (-4 square root 2 , 135°)
D) (4 square root 2 , 225°), (-4 square root 2 , 45°)

17. The graph of a limacon curve is given. Without using your graphing calculator, determine which equation is correct for the graph.
a circular graph with an inner loop on the left


[-5, 5] by [-5, 5] (5 points)

A) r = 3 + 2 cos θ
B) r = 2 + 3 cos θ
C) r = 2 + 2 cos θ
D) r = 4 + cos θ

18. Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.
r = -2 + 3 cos θ

A) No symmetry
B) y-axis only
C) x-axis only
D) Origin only

19. A railroad tunnel is shaped like a semiellipse, as shown below.
A semiellipse is shown on the coordinate plane with vertices on the x axis and one point of intersection with the positive y axis.

The height of the tunnel at the center is 54 ft, and the vertical clearance must be 18 ft at a point 8 ft from the center. Find an equation for the ellipse.


20. Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.
r = 2 cos 3θ

Answer 1

11. The technique to answering this problem quickly is by examining the choices. We know that the denominator for y and x should be equal to the square of the foci (from the fact that $c^{2}=a^{2}+b^{2}$). This would mean that the denominators should add up to 81. From the choices, only two satisfy this and those are options A and D. We know it should be D since the smaller denominator always has to be in the first term.

ANSWER: D. y squared over 36 minus x squared over 45 = 1.

12. You can also quickly identify the equation of the hyperbola given the vertices and the asymptotes. The square root of the denominator of the FIRST TERM in the equation is the numerator of the asymptote while the square root of the denominator of the SECOND TERM is the denominator of the asymptote. HOWEVER, we have to consider that the vertices are at (0,4) and (0,-4) so the asymptote must be in the lowest term. Considering the vertices, we can arrive at the asymptote $y=+- \frac{4}{16}x$. This means that the first term will have a denominator of 16 while the second term will have a denominator of 256. This equation is option B.

ANSWER: B. y squared over 16 minus x squared over 256 = 1.

13. To eliminate the parameter, t, we just need to equate both equations such that t is equal.

$t=x+3$
$t=\sqrt{y-5}$

$x+3=\sqrt{y-5}$
$x^{2}+6x+9=y-5$
$y=x^{2}+6x+14$

ANSWER: A. $y=x^{2}+6x+14$

14. To find the x coordinate we just need to multiply r and cos θ while for the y coordinate we would need to multiply r and sin θ.

$x=(3)[cos(\frac{2\pi}{3})]=-\frac{3}{2}$
$y=(3)[sin(\frac{2\pi}{3})]=\frac{3\sqrt{3}}{2}$

ANSWER: A. ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2.

15. The same polar coordinate as point P will be arrived if: (1) a full rotation of 2nπ is performed or (2) a rotation of (2n+1)π is performed and r is negated. Among the choices, we can see exactly one option with coordinates that follow these two rules, and it is choice D.

ANSWER: D. (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n+1)π)

16. To convert the coordinate (4, 4) to polar, we just let x be rcosθ and y be rsinθ. We can get r by solving $r^{2}=x^{2}+y^{2}$ while we can get θ by solving for $\theta=arctan(\frac{y}{x})$.

$r^{2}=(4)^{2}+(4)^{2}$
$r=4sqrt(2)$

$\theta=arctan(\frac{4}{4})=45degrees$

Therefore, one pair of polar coordinates would be (4 square root of 2, 45 degrees) and another one would be (-4 square root of 2, 225 degrees) [note the rule we stated in number 15].

ANSWER: B. (4 square root of 2, 45 degrees), (-4 square root of 2, 225 degrees)

17. Based on your definition that the graph is "circular" with an inner loop on the left, we can only deduce that the limacon with the form r = a + bcosθ has a value of b that is greater than a. Looking at the choices, we only have one option following this criteria, thus we can be sure that it is the correct answer.

ANSWER: B. r = 2 + 3cos θ

18. To test for symmetry about the x-axis, we replace the variables r and θ with r and -θ respectively or -r and π-θ. The equation will be symmetric if it will be unchanged (i.e. the same points will still satisfy the new equation).

$r=-2+3cos\theta$
$r=-2+3cos(-\theta)$
$-r=-2+3cos(\pi-\theta)$

If you examine closely, the same set of points will satisfy the equation above. Therefore, the equation is symmetric about the x axis.

For symmetry about the y-axis, we replace r and θ with -r and -θ.

$r=-2+3cos\theta$
$-r=-2+3cos(-\theta)$

Unfortunately, the equation was changed upon substitution therefore we know that it is not symmetric about the y-axis.

For symmetry about the origin, we just replace r and θ with -r and θ respectively.

$r=-2+3cos\theta$
$-r=-2+3cos\theta$

These two equations are not also similar so no symmetry about the origin is exhibited. 

ANSWER: C. x-axis only

19. Let's consider the general form of the ellipse: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$.

From the problem we know that a is equal to 54 since we are given the fact that the height of the tunnel is 54ft. To find b, we use the fact that the point (8,18) is a point on the ellipse as stated in the problem.

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{54^{2}}=1$
$\frac{8^{2}}{b^{2}}+\frac{18^{2}}{54^{2}}=1$
$\frac{8^{2}}{b^{2}}+\frac{1}{9}=1$
$64=(\frac{8}{9})b^{2}$
$b^{2}=72$

ANSWER: $\frac{x^{2}}{72}+\frac{y^{2}}{2916}=1$

20. For this item, we follow similar rules stated in item #18. To test for symmetry we need to examine if the equation will remain unchanged after performing substitutions. 

$r=2cos3\theta$

x-axis:
$r=2cos(-3\theta)$
$-r=2cos[3(\pi-\theta)]$

y-axis:
$-r=2cos(-3\theta)$

origin:
$-r=2cos3\theta$

If you examine the equations, you'll see that it is only symmetric at the x-axis.

ANSWER: The equation is symmetric about the x-axis.