11. Ans: (<span>
D)
</span>Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:
![\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B%28y-k%29%5E2%7D%7Ba%5E2%7D%20-%20%20%5Cfrac%7B%28x-h%29%5E2%7D%7Bb%5E2%7D%20%3D%201%20%20)
Since (h,k) = (0,0)
Therefore, the above equation becomes,
![\frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B%28y%29%5E2%7D%7Ba%5E2%7D%20-%20%5Cfrac%7B%28x%29%5E2%7D%7Bb%5E2%7D%20%3D%201%20)
Now the distance between the vertices is:
2a = 12
=> a = 6
And the distance between the foci is:
2c = 18
=> c = 9
Since,
![c^2 = a^2 + b^2](https://tex.z-dn.net/?f=c%5E2%20%3D%20a%5E2%20%2B%20b%5E2)
=>
![b^2 = 45](https://tex.z-dn.net/?f=b%5E2%20%3D%2045)
Hence, the equation becomes,
(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):
![\frac{y^2}{16} - \frac{x^2}{b^2} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7By%5E2%7D%7B16%7D%20-%20%20%5Cfrac%7Bx%5E2%7D%7Bb%5E2%7D%20%3D%201%20%20)
-- (X)
=> ![y^2 = ( \frac{16}{b^2})*(b^2 + x^2)](https://tex.z-dn.net/?f=y%5E2%20%3D%20%28%20%5Cfrac%7B16%7D%7Bb%5E2%7D%29%2A%28b%5E2%20%2B%20x%5E2%29%20)
=> y = ±
![( \frac{4}{b} ).x](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B4%7D%7Bb%7D%20%29.x)
--- (A)
Since asymptotes at y = ±
![( \frac{1}{4} ).x](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B1%7D%7B4%7D%20%29.x)
. --- (B)
Compare (A) and (B), you would get,
![\frac{4}{b} = \frac{1}{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7Bb%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
=> b=16
The equation (X) would become:
![\frac{y^2}{16} - \frac{x^2}{256} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7By%5E2%7D%7B16%7D%20-%20%5Cfrac%7Bx%5E2%7D%7B256%7D%20%3D%201%20)
(
Option-B)
13. Ans: (A) ![y = x^{2} + 6x + 14](https://tex.z-dn.net/?f=y%20%3D%20x%5E%7B2%7D%20%2B%206x%20%2B%2014)
<span>Equations given:
</span><span>x = t - 3 --- (equation-1)
y =
![t^{2}](https://tex.z-dn.net/?f=t%5E%7B2%7D)
+ 5 --- (equation-2)
From equation-1,
t = x + 3
Put the value of t in (equation-2),
</span>
![y = (x+3)^{2} + 5](https://tex.z-dn.net/?f=y%20%3D%20%28x%2B3%29%5E%7B2%7D%20%2B%205)
![y = x^2 + 9 + 6x + 5](https://tex.z-dn.net/?f=y%20%3D%20x%5E2%20%2B%209%20%2B%206x%20%2B%205)
![y = x^2 + 6x + 14](https://tex.z-dn.net/?f=y%20%3D%20x%5E2%20%2B%206x%20%2B%2014)
<span>
Hence, the correct option is
(A)
14. Ans: (A) </span>
The polar coordinates given:
![(3, \frac{2 \pi }{3} )](https://tex.z-dn.net/?f=%283%2C%20%20%5Cfrac%7B2%20%5Cpi%20%7D%7B3%7D%20%29)
= (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
![\frac{2 \pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Cpi%20%7D%7B3%7D)
= 120°
=>
x =
![- \frac{3}{2}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B3%7D%7B2%7D%20)
y = (3) * sin(120°);
=>
y =
![\frac{3 \sqrt{3} }{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%20%5Csqrt%7B3%7D%20%7D%7B2%7D%20)
Ans: (x,y) = ![(- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})](https://tex.z-dn.net/?f=%28-%20%5Cfrac%7B3%7D%7B2%7D%20%2C%5Cfrac%7B3%20%5Csqrt%7B3%7D%20%7D%7B2%7D%29)
(
Option A)
15. Ans: (D)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive):
(r
, θ + 2n
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
) where, n = integer
2) When r < 0(meaning negative):
(-r
, θ + (2n+1)
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
) 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) =
![\frac{- \pi }{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-%20%5Cpi%20%7D%7B6%7D%20)
When r = +1(r>0):
(1,
![\frac{- \pi }{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-%20%5Cpi%20%7D%7B6%7D%20)
+ 2n
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
)
When r = -1(r<0):
(-1,
![\frac{- \pi }{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-%20%5Cpi%20%7D%7B6%7D%20)
+ (2n+1)
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
)
Therefore, the correct option is
(D): <span>(1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)
</span>
16. Ans: (<span>
B)</span>In polar coordinates,
![r = \sqrt{x^{2} + y^{2}}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%7Bx%5E%7B2%7D%20%2B%20y%5E%7B2%7D%7D%20)
Since x = 4, y=4; therefore,
![r = \sqrt{16 + 16} = 4 \sqrt{2}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%7B16%20%2B%2016%7D%20%20%3D%204%20%5Csqrt%7B2%7D%20)
To find the angle,
tanθ = y/x = 4/4 = 1
=> θ =
45° (when
![r =4 \sqrt{2}](https://tex.z-dn.net/?f=r%20%3D4%20%5Csqrt%7B2%7D%20)
)
If r = -
![r =4 \sqrt{2}](https://tex.z-dn.net/?f=r%20%3D4%20%5Csqrt%7B2%7D%20)
, 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:
![\frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D%20%2B%20%20%5Cfrac%7By_%7B2%7D%7D%7Bb_%7B2%7D%7D%20%3D%201%20%20)
-- (A)
Since, x = 8f,
y = 18ft,
b = 54ft,
![a^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D)
= ?
Plug-in the values in equation (A),
(A)=>
![\frac{64}{a^{2}} + \frac{324}{2916} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B64%7D%7Ba%5E%7B2%7D%7D%20%2B%20%20%5Cfrac%7B324%7D%7B2916%7D%20%3D%201%20%20)
=>
![a^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D)
= 72
Therefore, the equation becomes,
Ans: ![\frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B72%7D%20%2B%20%5Cfrac%7By_%7B2%7D%7D%7B2916%7D%20%3D%201%20)
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!