# Math Help - need help with 10 questions

1. ## need help with 10 questions

1)
Give the focus, directrix, and axis of the parabola.
x = 6y2

2)
Find the foci and vertices of the ellipse.
25
x2 + 4y2 16 = 0

3)
Find the asymptotes of the hyperbola.
x^2/25-y^2/4=1

4)
Find
θ so that rotating axes the through an angle of θ produces a new equation containing no xy-term.
31x2 + 10sqrt(3xy) + 21y2 −144 = 0

5)
Use the angle found in problem 4
to rotate the conic and eliminate the xy term. Write the equation of
rotated conic.
31x2 + 10sqrt(3xy) + 21y2 −144 = 0

6)

Identify the conic produced and the angle of rotation from problem 4
31x2 + 10sqrt(3xy) + 21y2 −144 = 0

7)
Find the rectangular coordinates for the point.
(
3,2pi/3)

8)
Change the equation to polar coordinates.
xy = 1

and
x2 + y2 8y = 0

9)
Change the equation to rectangular coordinates.

r = 2
and
r=sin(theta)

10)
Find a rectangular equation for the plane curve defined by the parametric equations.
x = 4sin θ, y = cos θ

and
x
= tan(t), y = sec(t)

If you're doing Rotations, you must be familiar with certain formulas.

4. Find θ so that rotating axes the through an angle of θ
produces a new equation containing no xy-term.
. . 31x² + (10√3)xy + 21y² − 144 .= .0
The general quadratic equation is: .Ax² + Bxy + Cy² + Dx + Ey + F .= .0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B
The angle of rotation θ is given by: .tan(2θ) .= .------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C

. . . . . . . . . . . . . . . . . . . _
We have: .A = 31, B = 10√3, C = 21, D = 0, E = 0, F = -144
. . - . . - . . . . . . . . . . . _
. . . . . . . . . . . . . . .10√3 . - - . _
Hence: . tan2θ .= .--------- .= .√3
. . . . . . . . . . . . . .31 - 21

Then: . .= .π/3 . . . . θ .= .π/6

5) Use the angle found in problem 4 to rotate the conic and eliminate the xy term.
Write the equation of rotated conic.
To eliminate the xy-term, we use these substitutions:
. . x' .= .x·cosθ - y·sinθ
. . y' .= .x·sinθ + y·cosθ

With θ = π/6, we have:
. . x .= .½√3·x - ½·y
. . y .= .½·x + ½√3·y

Substitute:

31(½√3x - ½y)² + 10√3(½√3x - ½y)(½x + ½√3y) + 21(½x + ½√3y)² - 144 .= .0

This simplifies to: .36x² + 16y² .= .144

6) Identify the conic produced and the angle of rotation from problem 4
. . . . . . . . . . . . . . . . . . . .
The conic is an ellipse: . --- + --- .= .1
. . . . . . . . . . . . . . . . . .4 . . .9

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could have identitied the conic long ago.

The equation has a discriminant: .d .= .B² - 4AC . .
(Look familiar?)
. . If d = 0: parabola
. . If d > 0: hyperbola
. . If d < 0: ellipse

. . . . . . . . . . . . . . . . . . . _
We have: .A = 31, B = 10√3, C = 21
. . - . . . - . . . - . . . . _
Therefore: .d .= .(10√3)² - 4(31)(21) .= .-2304 . . . . ellipse

3. Originally Posted by Soroban

If you're doing Rotations, you must be familiar with certain formulas.

The general quadratic equation is: .Ax² + Bxy + Cy² + Dx + Ey + F .= .0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B
The angle of rotation θ is given by: .tan(2θ) .= .------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C

. . . . . . . . . . . . . . . . . . . _
We have: .A = 31, B = 10√3, C = 21, D = 0, E = 0, F = -144
. . - . . - . . . . . . . . . . . _
. . . . . . . . . . . . . . .10√3 . - - . _
Hence: . tan2θ .= .--------- .= .√3
. . . . . . . . . . . . . .31 - 21

Then: . .= .π/3 . . . . θ .= .π/6

To eliminate the xy-term, we use these substitutions:
. . x' .= .x·cosθ - y·sinθ
. . y' .= .x·sinθ + y·cosθ

With θ = π/6, we have:
. . x .= .½√3·x - ½·y
. . y .= .½·x + ½√3·y

Substitute:

31(½√3x - ½y)² + 10√3(½√3x - ½y)(½x + ½√3y) + 21(½x + ½√3y)² - 144 .= .0

This simplifies to: .36x² + 16y² .= .144

. . . . . . . . . . . . . . . . . . . .
The conic is an ellipse: . --- + --- .= .1
. . . . . . . . . . . . . . . . . .4 . . .9

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could have identitied the conic long ago.

The equation has a discriminant: .d .= .B² - 4AC . .
(Look familiar?)
. . If d = 0: parabola
. . If d > 0: hyperbola
. . If d < 0: ellipse

. . . . . . . . . . . . . . . . . . . _
We have: .A = 31, B = 10√3, C = 21
. . - . . . - . . . - . . . . _
Therefore: .d .= .(10√3)² - 4(31)(21) .= .-2304 . . . . ellipse

4. i got up to the simplifying part just having a hard time simplifying heh
thanks alot

well it seems i got everything besides 7-10 if you can help me with those thatll be great im working on it, its just taking a while sigh

i got up to the simplifying part just having a hard time simplifying heh
thanks alot

well it seems i got everything besides 7-10 if you can help me with those thatll be great im working on it, its just taking a while sigh
See my first post here http://www.mathhelpforum.com/math-he...ordinates.html for the rules

I'll do one of your questions as an example, try the rest and tell me if you get it

7) (3, 2pi/3) to rectangular coordinates

=> (r, theta) = (3, 2pi/3)

x = rcos(theta) = 3cos(2pi/3) = -3/2
y = rsin(theta) = 3sin(2pi/3) = 3sqrt(3)/2

so the rectangular coordinates are: (-3/2 , 3sqrt(3)/2)

now try the other questions of this type (8 and 9)

6. 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

10)
Find a rectangular equation for the plane curve defined by the parametric equations.
x = 4sin θ, y = cos θ

we start by solving for the paramentric variable in terms of x, then we replace the parametric variable with the expression in y

x = 4sin(theta)
=> sin(theta) = x/4 .............remember this, we'll get back to it
=> theta = arcsin(x/4)

now y = cos(theta)
=> y = cos(arcsin(x/4))
now recall that arcisn(x/4) = theta and sin(theta) = x/4
so we draw a right triangle with one acute angle being theta, the length of the side opposite to the angle is x and the hypotenuse is length 4. by pythagoras, the adjacent side is sqrt(16 - x^2), from this triangle we see that:

y = cos(arcsin(x/4)) = cos(theta) = sqrt(16 - x^2)/4
=> 4y = sqrt(16 - x^2)
=> 16y^2 = 16 - x^2
=> x^2 + 16y^2 = 16
=> (x^2)/16 + y^2 = 1
this is an ellipse. now try the other part of this question

8. k i used a different method to solving number 10 buti got it

k i used a different method to solving number 10 buti got it
what method was that?

10. well what i did was x=tan(t) y=sec(t)
1+tan^2=sec^2
1+x^2=y^2
y^2-x^2=1
hyperbola
similarily x=4sin(@)
x/4=sin(@)
sin^2+cos^2=1
(x/4)^2+y^2=1 same answer though can you help me with
number 8 and second part of number 9

well what i did was x=tan(t) y=sec(t)
1+tan^2=sec^2
1+x^2=y^2
y^2-x^2=1
hyperbola
similarily x=4sin(@)
x/4=sin(@)
sin^2+cos^2=1
(x/4)^2+y^2=1 same answer though can you help me with
number 8 and second part of number 9
that's nice! i never thought of using identities. i should remember that technique for the future

8)
Change the equation to polar coordinates.
xy = 1

and
x2 + y2 − 8y = 0

remember, x = rcos(theta) and y = rsin(theta)
so xy = 1
=> rcos(theta)*rsin(theta) = 1
=> r^2 * sin(theta)cos(theta) = 1
=> r^2 = 1/(sin(theta)cos(theta))
=> r^2 = sec(theta)csc(theta)

9)
Change the equation to rectangular coordinates.

r=sin(theta)

r = sin(theta)
=> r^2 = rsin(theta)
=> x^2 + y^2 = y
=> x^2 + y^2 - y = 0
=> x^2 + y^2 - y + (-1/2)^2 = (-1/2)^2
=> x^2 + (y - 1/2)^2 = (1/2)^2
this is a circle with center (0, 1/2) and radius 1/2

13. 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

14. and for 9 how did you get one half can you walk me through that problem again, it sort of doesnt make sense