# Thread: need help with 10 questions

question number 8 doesnt really make sense isnt it suppose to be polar coordinates? what i tried to do for xy=1 is uhmm x and y has to rather equal pi/4 or 5pi/4(cuz its int he 3rd quardnant) from there i did tan(@)=1 @=pi/4 or 5pi/4 x=rcos x=1 therefore 1=rcos(@) 1=r(1/sqrt2) r=+-sqrt2 and for sin i got pi/4+npi
yeah, the question said polar coordinates, but i believe it should have said "polar equation" since we were not given coordinates, we were given an equation. therefore, i just replaced x with rcos(theta) and y with rsin(theta) and simplified

2. that makes sense, hey can you help me with second part of number 8 that one isnt working for me i dont know why but i need the coordinates and can you check my answers for 1-4 that i posted earlier thatll be great
thanks

and

8)
Change the equation to polar coordinates.
x2 + y2 8y = 0
again, just replace x with rcos(theta) and y with rsin(theta) and i'm tired of typing theta, i'll use t from now on

x^2 + y^2 - 8y = 0
=> (rcos(t))^2 + (rsin(t))^2 - 8(rsin(t)) = 0
=> r^2*cos^2(t) + r^2*sin^2(t) - 8rsin(t) = 0
=> r^2*(cos^2(t) + sin^2(t)) - 8rsin(t) = 0
=> r^2 - 8rsin(t) = 0
and i'd leave it there

k can you guys check if im doing something wrong
for question 1
x=6y^2
y^2=x/6
y2=4px
4px=1/6 therefore p=1/24
therefore focus h+p=(1/24,0)
axis= 0 since k=0
directrix=h-p=-1/24

question 2
25x2+4y2-16=0
25x2/4^2+y2/2^2=1

a2+b1=c2 16+4=c2 2sqrt5=c therefore foci (+-2sqrt5,0) verticies(0,+-2)

question 3
x2/25-y2/4=1
y2=4+(4x2/25)
y=+-(2sqrt(x2+25))/5

and i dont know if that polar coordinate link is applicable with 8 im still working on it
can check if these answers are right before i submit it?
thatll be great thanks

5. i find questions like these so annoying i don't know why. whenever i see "find the directrix and focus of a parabola", i cringe. anyway...

k can you guys check if im doing something wrong
for question 1
x=6y^2
y^2=x/6
y2=4px
4px=1/6 therefore p=1/24
therefore focus h+p=(1/24,0)
axis= 0 since k=0
directrix=h-p=-1/24
this looks fine

question 2
25x2+4y2-16=0
25x2/4^2+y2/2^2=1

a2+b1=c2 16+4=c2 2sqrt5=c therefore foci (+-2sqrt5,0) verticies(0,+-2)
for the foci i got (0, +/- sqrt(84)/5) ........check my calculations. for (x^2)/(b^2) + (y^2)/(a^2) = 1 where a >= b > 0 the foci is (0, +/- c) where c^2 = a^2 - b^2

question 3
x2/25-y2/4=1
y2=4+(4x2/25)
y=+-(2sqrt(x2+25))/5
for asymptote:

y = +/- (b/a)x = +/- (2/5)x

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