Question

PLEASE HELP

Answer 1

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:

$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

Since (h,k) = (0,0)
Therefore, the above equation becomes,
$\frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1$

Now the distance between the vertices is:
2a = 12
=> a = 6

And the distance between the foci is:
2c = 18
=> c = 9

Since,
$c^2 = a^2 + b^2$

=> $b^2 = 45$

Hence, the equation becomes,
$\frac{(y)^2}{36} - \frac{(x)^2}{45} = 1$  (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):
$\frac{y^2}{16} - \frac{x^2}{b^2} = 1$ -- (X)

=> $y^2 = ( \frac{16}{b^2})*(b^2 + x^2)$

=> y = ± $( \frac{4}{b} ).x$ --- (A) 
Since asymptotes at y = ± $( \frac{1}{4} ).x$. --- (B)
Compare (A) and (B), you would get,
$\frac{4}{b} = \frac{1}{4}$

=> b=16

The equation (X) would become:
$\frac{y^2}{16} - \frac{x^2}{256} = 1$ (Option-B)


13. Ans: (A) $y = x^{2} + 6x + 14$
Equations given:
x = t - 3 --- (equation-1)
y = $t^{2}$ + 5 --- (equation-2)

From equation-1,
t = x + 3

Put the value of t  in (equation-2),
$y = (x+3)^{2} + 5$
$y = x^2 + 9 + 6x + 5$
$y = x^2 + 6x + 14$

Hence, the correct option is (A)

14. Ans: (A) 

The polar coordinates given: $(3, \frac{2 \pi }{3} )$ = (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 $\frac{2 \pi }{3}$ = 120°
=> x = $- \frac{3}{2}$

y = (3) * sin(120°);
=> y = $\frac{3 \sqrt{3} }{2}$

Ans: (x,y) = $(- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})$ (Option A)

15. Ans: (D)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive): (r, θ + 2n $\pi$) where, n = integer 
2) When r < 0(meaning negative): (-r, θ + (2n+1) $\pi$) 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) = $\frac{- \pi }{6}$

When r = +1(r>0):
(1, $\frac{- \pi }{6}$ + 2n$\pi$)

When r = -1(r<0):
(-1, $\frac{- \pi }{6}$ + (2n+1)$\pi$)

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,
$r = \sqrt{x^{2} + y^{2}}$

Since x = 4, y=4; therefore,
$r = \sqrt{16 + 16} = 4 \sqrt{2}$

To find the angle,
tanθ = y/x = 4/4 = 1

=> θ = 45° (when $r =4 \sqrt{2}$)
If r =  -$r =4 \sqrt{2}$, 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:

$\frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1$ -- (A)

Since, x = 8f,
y = 18ft,
b = 54ft,
$a^{2}$ = ? 

Plug-in the values in equation (A),
(A)=> $\frac{64}{a^{2}} + \frac{324}{2916} = 1$

=> $a^{2}$ = 72

Therefore, the equation becomes,
Ans: $\frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1$


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!