Question

PLEASE HELP

Answer 1

Question 1
We are given horizontal hyperbola in its conic form:
$\frac{(x-1)^2}{49}-\frac{(y+3)^2}{9}=1$
The general formula for hyperbola in conic form is:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
This is a horizontal hyperbola if we want to get vertical hyperbola we simply switch places of y and x:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1$
If we look at our parabola we can see that a=7, b=3, h=1 and k=-3.
We know that center of a parabola is at the point (h,k) and that vertices are at (h+a,k) and (h-a,k). 
We that mind we find:
$Center:(1,-3)\\ Vertices:(8,-3),(-6,-3)$
The answer is C. Please click this link to see that graph(https://www.desmos.com/calculator/g2pbnxs9ad)
Question 2
Standard form is the same as the conic form that we mention in part 1.
We are given:
$Vertices:(0,\pm 2)\\ Foci:(0,\pm 11)$
What we can notice right away is that this is a vertical hyperbola. 
We know that vertices are at:
$(h,k\pm a)$
And that foci are at:
$(h,k\pm c)$
We know that h=0. We can also conclude that k=0. We know that vertices are a point far away from the center. So our vertices are:
$k+a=2\\ k-a=-2$
We solve this system and we get that a=2 and k=0. That means that our center is at (0,0).
Our foci are given with these two equations:
$k+c=11\\ k-c=-11$
Since we know that k=0, c=11. We that in mind we can find b:
$c^2=a^2+b^2\\ b^2=c^2-a^2\\ b^2=117\\ b=\sqrt{117}$
Now we have all the information to write the equation for the parabola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{4}-\frac{x^2}{117}=1$
The answer is A. Link to the graph: https://www.desmos.com/calculator/gw6fnf8pg0 .
Question 3
We are given:
$Vertices:(0,\pm 8)\\ Asymptotes:y=\pm \frac{x}{2}$
We notice that this is a vertical hyperbola (look at the coordinates of vertices, they are changing along the y-axis).
We know that our vertices are given by the following formula:
$k+a=8\\ k-a=8$
From this, we can conclude that k=8, and a=8. We know that our center is at (0,0).
Asymptote for the vertical hyperbola is given by this formula:
$y=\pm \frac{a}{b}(x-h)+k$
For the horizontal hyperbola, you just swap a and b. 
We knot that our h and k are zeros, and if we look at our asymptotes we notice that:
$\frac{a}{b}=\frac{1}{2}$
We know that a=8 so we can find b from the above equation:
$\frac{8}{b}=\frac{1}{2}\\ b=16$. 
This is all we need to write down the equation of this hyperbola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{64}-\frac{x^2}{256}=1$
The answers is B. Link to the graph(https://www.desmos.com/calculator/tbecgjct6g).
Question 4
We are given these two equations:
$x=t-3\\ y=\frac{2}{t+5}$
To eliminate t, we will express t, from the first equation, in terms of x and then plug it back into the second equation:
$x=t-3\\ y=\frac{2}{t+5}\\ t=x+3\\ y=\frac{2}{x+3+5}\\ y=\frac{2}{5+8}$
The answer is D.
Question 5
We are given polar coordinates:
$(7,\frac{2\pi}{3})$
To conver polar cordinates in rectangular one with use the following formula:
$x=r\cos(\theta)\\ y=r\sin(\theta)$
In our case r=7 and $\theta =\frac{2\pi}{3}$.
Let us calculate the rectangular coordinates:
$x=7\cos(\frac{2\pi}{3})\\ y=7\sin(\frac{2\pi}{3})\\ x=\frac{-7}{2}\\ y=\frac{7\sqrt{3}}{2}}$
Coordinates are:
$(\frac{-7}{2},\frac{7\sqrt{3}}{2}})$
The answer is A.
Question 6
We are given a point P with the following coordinates:
$P(1,\frac{\pi}{3})$
What you should keep in mind is that for any point there are two pairs of polar coordinates, because you can reach the same point rotating your point vector clock-wise or counter-clockwise. If you are given a pair of polar coordinates this is the formula you can use to get the second pair:
$(r,\theta),(-r,\theta+\pi)$
Reason for this is because you rotate your pointing vector by 180 degrees. This means you are doing a reflection on the y=x line and therefore you end up with the same point if you use -r instead of r.With that in mind the second pair for our points is:
$(-1,\frac{\pi}{3}+\pi)\\ (-1,\frac{4\pi}{3})$
If we rotate our pointing vector by 360 degrees(2pi radians) we will always end up at the same point. In fact, if you any number of full circles you end with the same point.So, the final answer is:
$(-1,\frac{4\pi}{3}+2n\pi),(1,\frac{\pi}{3}+2n\pi)$
None of the answers you provided are correct. You can change r to -r and end up at the same point without changing the angle.