SOLUTION: Can someone show me how to do these two problems please? 1:Write an equation for the hyperbola where all points on the hyperbola are 68 units closer to one focus than the other.

Question 428547: Can someone show me how to do these two problems please?
1:Write an equation for the hyperbola where all points on the hyperbola are 68 units closer to one focus than the other. The foci are located at (0,0) and (250,0).
2: Write an equation for an ellipse with center (1,-3), vertices (1,2) and (1,-8) and co-vertices (4,-3) and (-2,-3)
Show me the steps please, I need to learn how to do this.
Thanks
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1:Write an equation for the hyperbola where all points on the hyperbola are 68 units closer to one focus than the other. The foci are located at (0,0) and (250,0).
..
Standard form of hyperbola with horizontal transverse axis(opens sideways):
(x-h)^2a^2-(y-k)^2/b^2=1
Standard form of hyperbola with Vertical transverse axis(opens up and down):
(y-k)^2a^2-(x-h)^2/b^2=1
(h,k)=(xy) coordinates of the center for both forms
..
2a=68 (points on hyperbola closer to one focus than the other)
a=34
a^2=1156
Center:(125,0)
Foci:(0,0) and (250,0)=c (given location of foci)
Tlhis shows transfer axis is horizontal at y=0
c=125
c^2=15625
c^2=a^2+b^2
b^2=c^2-a^2=15625-1156=14469
b=sqrt(14469)=120.29
..
Equation of Hyperbola:
(x-125)^2/1156-y^2/14469=1
Equations of asymptotes: (use standard form for a straight line, y=mx+b, with slopes +-b/a and a point on the line which is the center,(125.0)
y=(120.29/34)x-442
y=-(120.29/34)x+442
see the graph below which can serve as a check on the answer
..
y=(((((x-125)^2)/1156)-1)*14469)^.5

...
2: Write an equation for an ellipse with center (1,-3), vertices (1,2) and (1,-8) and co-vertices (4,-3) and (-2,-3)
Standard form of ellipse for horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1 (a>b)
Standard form of ellipse for vertical major axis: (y-k)^2/a^2+(x-h)^2/b^2=1 (a>b)
In both forms, (h,k) represent the (x,y) coordinates of the center.
..
Given data shows ellipse has a vertical major axis on the line,x=1, so it is of the second form described above.
Length of major axis =10=2a
a=5
a^2=25
Length of minor axis=6=2b
b=3
b^2=9
center:(1,-3) given
We now have all the information we need to write an equation for the ellipse.
(y+3)^2/25+(x-1)^2/9=1
see the graph below to check the answers
..
y=(25(1-(x-1)^2/9))^.5-3

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