SOLUTION: i need a lot of help with these questions i do not understand them at all..! pleas help.! 1.What is the distance between the following points? (-2 sq root 7,10) (4 sq rt 7,8)

Question 421697: i need a lot of help with these questions i do not understand them at all..! pleas help.!
1.What is the distance between the following points?
(-2 sq root 7,10) (4 sq rt 7,8)
2.What is the axis of symmetry of the parabola with the equation x � 4 = 1/4(y + 1)2
3.Find the foci of an ellipse with the equation 7(x � 2)2 + 3(y � 2)2 = 21.
4.Find the vertices of a hyperbola with the equation (x + 3)2 � 4(y � 2)2 = 4
5.Find the foci of a hyperbola with the equation 9y2 � 72y � 16x2 � 64x � 64 = 0
i need the last few to be in (a,b) form.. Thanks.!
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1.What is the distance between the following points?
(-2 sq root 7,10) (4 sq rt 7,8)
2.What is the axis of symmetry of the parabola with the equation
x � 4 = 1/4(y + 1)2
3.Find the foci of an ellipse with the equation 7(x � 2)2 + 3(y � 2)2 = 21.
4.Find the vertices of a hyperbola with the equation (x + 3)2 � 4(y � 2)2 = 4
5.Find the foci of a hyperbola with the equation 9y2 � 72y � 16x2 � 64x � 64 = 0 ..
1.The formula for finding distance between two points:
d=sqrt[(x1-x2)^2+(y1-y2)^2]
for given points,
distance=sqrt[(4sqrt7-(-2sqrt7))^2+(8-10)^2]
=sqrt[(6sqrt7)^2+(-2)^2]
=sqrt[252+4]=sqrt256=16 (ans)
..
2.x�4=1/4(y+1)2
This can be rewritten:
x-4=1/2(y+1)^2
or x=1/2(y+1)^2+4
This is now written in standard form of a parabola, x=A(y-k)^2+h, with (h,k) being the (x,y) coordinates of the parabola.
For the given equation, this is a parabola that opens sideways to the right, with the vertex at (4,-1). The axis of symmetry is y=-1
The graph below will help you understand what some of the numbers above represent:
..
y=-1+-(2x-8)^.5

..
3. 7(x�2)^2+3(y�2)^2=21
Standard form for an ellipse:(x-h)^2/a^2+(y-k)^2/b^2
or(y-k)^2/a^2+(x-h)^2/b^2,a>b
for given equation, divide by 21 to write it in standard form,
(x�2)^2/3+(y�2)^2/7=1 (Since the denominator of the y^2-term is greater than
the denominator of the x^2-term, the given equation is of the second standard
form listed above.)
(y�2)^2/7+(x�2)^2/3=1
This an ellipse with center at (2,2) with vertical major axis.
a^2=7
a=sqrt7
b^2=3
b=sqrt3
c^2=a^2-b^2=7-3=4
c=sqrt4=2
foci=+-c from the center on the vertical major axis
=(2,4) and (2,0)
The graph of the given ellipse below may help you understand some of the
numbers I came up with:
..
y=2+-(7-(7/3)(x-2)^2)^.5

..
4.Find the vertices of a hyperbola with the equation (x+3)^2�4(y�2)^2=4
(x+3)^2�4(y�2)^2=4
Standard form of a hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1. Notice this form is almost as same as that of an ellipse except a hyperbola has a negative second term.
Divide given equation by 4,
(x+3)^2/4�(y�2)^2/1=1
This is a hyperbola with center at (-3,2) and it has a horizontal transverse axis at y=2, that is, it opens sideways.
a^2=4
a=2
b^2=1
b=1
c^2=a^2+b^2
c=sqrt(4+1)=sqrt(5)=2.24
vertices on the transverse axis=-3+-a=(-5,2) and (-1,2)
Foci on the transverse axis=-3+-c=-3+-2.24=(-5.24,2) and (.76,2)
The graph of the given hyperbola below could serve as a check on some of the numbers like the center, vertices and foci.
..
y=2+-((x+3)^2/4-1)^.5

..
5.Find the foci of a hyperbola with the equation 9y2�72y�16x2�64x�64=0
9y2�72y�16x2�64x�64=0
you need to complete the square before you can see it in standard form:
9(y2�8y+16)�16(x2+4x+4)=64+144-64=144
9(y-4)^2-16(x+2)^2=144
Divide by 144
(y-4)^2/16-(x+2)^2/9=1
This is a hyperbola with center at (-2,4) with a vertical transverse axis,x=-2, that is it opens up and down.
a^2=16
a=4
b^2=9
b=3
c=sqrt(a^2+b^2)=sqrt(25)=5
On the vertical transverse axis,
foci:(-2,4+-c)=(-2,4+-5)=(-2,9) and (-2,-1)
The graph below can serve as a check on the center and foci.
..
y=4+-(16(1+(x+2)^2/9))^.5

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