11. Ans: (D)
Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:
[tex] \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 [/tex]
Since (h,k) = (0,0)
Therefore, the above equation becomes,
[tex] \frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1 [/tex]
Now the distance between the vertices is:
2a = 12
=> a = 6
And the distance between the foci is:
2c = 18
=> c = 9
Since,
[tex]c^2 = a^2 + b^2[/tex]
=> [tex]b^2 = 45[/tex]
Hence, the equation becomes,
[tex]\frac{(y)^2}{36} - \frac{(x)^2}{45} = 1[/tex] (Option D:y squared over 36 minus x squared over 45 = 1)
12. Ans: (B)
The hyperbola's standard form is(as it is a vertical):
[tex] \frac{y^2}{16} - \frac{x^2}{b^2} = 1 [/tex] -- (X)
=> [tex]y^2 = ( \frac{16}{b^2})*(b^2 + x^2) [/tex]
=> y = ± [tex]( \frac{4}{b} ).x[/tex] --- (A)
Since asymptotes at y = ± [tex]( \frac{1}{4} ).x[/tex]. --- (B)
Compare (A) and (B), you would get,
[tex] \frac{4}{b} = \frac{1}{4}[/tex]
=> b=16
The equation (X) would become:
[tex] \frac{y^2}{16} - \frac{x^2}{256} = 1 [/tex] (Option-B)
13. Ans: (A) [tex]y = x^{2} + 6x + 14[/tex]
Equations given:
x = t - 3 --- (equation-1)
y = [tex]t^{2}[/tex] + 5 --- (equation-2)
From equation-1,
t = x + 3
Put the value of t in (equation-2),
[tex]y = (x+3)^{2} + 5[/tex]
[tex]y = x^2 + 9 + 6x + 5[/tex]
[tex]y = x^2 + 6x + 14[/tex]
Hence, the correct option is (A)
14. Ans: (A)
The polar coordinates given: [tex](3, \frac{2 \pi }{3} )[/tex] = (r, θ)
Since,
x = r*cosθ,
y = r*sinθ
Plug-in the values of r, and θ in the above equations:
x = (3) * cos(120°); since [tex]\frac{2 \pi }{3}[/tex] = 120°
=> x = [tex]- \frac{3}{2} [/tex]
y = (3) * sin(120°);
=> y = [tex] \frac{3 \sqrt{3} }{2} [/tex]
Ans: (x,y) = [tex](- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})[/tex] (Option A)
15. Ans: (D)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive): (r, θ + 2n [tex] \pi [/tex]) where, n = integer
2) When r < 0(meaning negative): (-r, θ + (2n+1) [tex] \pi [/tex]) where, n = integer
Since r is not mentioned in the question, but in options every r slot has the value r=1, therefore, I would take r = +1, -1(plus minus 1)
θ(given) = [tex] \frac{- \pi }{6} [/tex]
When r = +1(r>0):
(1, [tex] \frac{- \pi }{6} [/tex] + 2n[tex] \pi [/tex])
When r = -1(r<0):
(-1, [tex] \frac{- \pi }{6} [/tex] + (2n+1)[tex] \pi [/tex])
Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)
16. Ans: (B)
In polar coordinates,
[tex]r = \sqrt{x^{2} + y^{2}} [/tex]
Since x = 4, y=4; therefore,
[tex]r = \sqrt{16 + 16} = 4 \sqrt{2} [/tex]
To find the angle,
tanθ = y/x = 4/4 = 1
=> θ = 45° (when [tex]r =4 \sqrt{2} [/tex])
If r = -[tex]r =4 \sqrt{2} [/tex], then,
θ = 45° + 180° = 225°
Therefore, the correct option is (B) (4 square root 2 , 45°), (-4 square root 2 , 225°)
17. Ans: (B)
(Question-17 missing Image is attached below) The general form of the limacon curve is:
r = b + a cosθ
If b < a, the curve would have inner loop. As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (B) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.
18. Ans: (C)
Let's find out!
1. If we replace θ with -θ, we would get:
r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2 + 3*cos(θ)
Same as the original, therefore, graph is symmetric to x-axis.
2. If we replace r with -r, we would get:
-r = -2 + 3*cos(θ )
r = 2 - 3*cos(θ)
NOT same as original, therefore, graph is NOT symmetric to its origin.
3. If we replace θ with -θ and r with -r, we would get:
-r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = 2 - 3*cos(3θ)
NOT same as original, therefore, graph is NOT symmetric to y-axis.
Ans: The graph is symmetric to: x-axis only!
19. (Image is attached below) As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:
[tex] \frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1 [/tex] -- (A)
Since, x = 8f,
y = 18ft,
b = 54ft,
[tex]a^{2}[/tex] = ?
Plug-in the values in equation (A),
(A)=> [tex] \frac{64}{a^{2}} + \frac{324}{2916} = 1 [/tex]
=> [tex]a^{2}[/tex] = 72
Therefore, the equation becomes,
Ans: [tex] \frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1 [/tex]
20. Ans: x-axis only
Let's find out!
1. If we replace θ with -θ, we would get:
r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = +2*cos(3θ) = Same as original
Therefore, graph is symmetric to x-axis.
2. If we replace r with -r, we would get:
-r = 2*cos(3θ )
r = -2*cos(3θ) = Not same
3. If we replace θ with -θ and r with -r, we would get:
-r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2*cos(3θ) = Not Same
Ans: The graph is symmetric to: x-axis only!
