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Question 459005: Please note that graphing is not allowed in solving this.
If (x^2/49)+(y^2/4)=1 and x^2=7y, then what are the values of x and y?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Please note that graphing is not allowed in solving this.
If (x^2/49)+(y^2/4)=1 and x^2=7y, then what are the values of x and y
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(x^2/49)+(y^2/4)=1
x^2=7y
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7y/49)+(y^2/4)=1
y/7+y^2/4=1
4y+7y^2=28
7y^2+4y-28=0
solve by quadratic formula
a=7, b=4, c=-28
y=[-4±sqrt(4^2-4*7*-28)]2*7
y=[-4±√(800)]14
y=(-4±28.28)/14
y=-2.31 (reject, y≥0)
or
y=1.73
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(x^2/49)+(y^2/4)=1
for y=1.73
(x^2/49)+(3/4)=1
x^2+36.75=49
x^2=12.25
x=±√12.25
x=±3.5
ans:
Points of intersection: (3.5,1.73) & (-3.5,1.73)
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check:
(x^2/49)+(y^2/4)=1
3.5^2/49+1.73^2/4
.25+.75..=1
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