Question 599465: What is the center, vertices, foci, and asymptotes of the equation 4y^2-25x^2+100x-8y-196=0?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the center, vertices, foci, and asymptotes of the equation
4y^2-25x^2+100x-8y-196=0?
complete the squares
4(y^2-2y+1)-25(x^2-4x+4)=196+4-100
4(y-1)^2-25(x-2)^2=100
divide by 100
(y-1)^2/25-(x-2)^2/4=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2, (h,k)=(x,y) coordinates of center
For given equation:
center: (2,1)
a^2=25
a=5
Vertices: (2,1±a)=(2,1±5)=(2,-4) and (2,6)
..
b^2=4
b=2
..
foci:
c^2=a^2+b^2=25+4=29
c=√29≈5.4
foci: (2,1±c)=(2,1±5.4)=(2,-4.4) and (2,6.4)
..
asymptotes:
slope of asymptotes=±a/b=±5/2
Asymptotes are straight lines that intersect at center (2,1)
Their standard form: y=mx+b, m=slope, b=y-intercept
..
Equation for asymptote with negative slope:
y=-5x/2+b
using coordinates of center to find b
1=-5*2/2+b
b=6
equation: y=-5x/2+6
..
Equation for asymptote with positive slope:
y=5x/2+b
using coordinates of center to find b
1=5*2/2+b
b=-4
equation: y=5x/2-4
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