SOLUTION: I need help putting 25x^2+16y^2+150x=160y-225 in standard form and then finding the a^2 value.
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Question 176404: I need help putting 25x^2+16y^2+150x=160y-225 in standard form and then finding the a^2 value.
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I need help putting 25x^2+16y^2+150x=160y-225 in standard form and then finding the a^2 value.
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Complete the square on the x-terms and on the y-terms:
25(x^2 + 6x + ?) + 16(y^2 - 10y + ?) = -225
25(x^2 + 6x + 9) + 16(y^2 - 10y + 25) = -225 + 25*9 + 16*25
25(x+3)^2 + 16(y-5)^2 = 400
[(x+3)^2]/16 + [(y-5)^2]/25 = 1
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a^2 = 25
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Cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation.
Subtract 160y from both sides.
Group like terms.
Factor 25 from the first group (to make the coefficient equal to 1)
Factor 16 from the second group (to make the coefficient equal to 1)
Take half of the "x" coefficient 6 to get 3. Square 3 to get 9. Add AND subtract this value inside the first parenthesis:
Add AND subtract 9 in the first parenthesis.
Factor to get
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Take half of the "y" coefficient -10 to get -5. Square -5 to get 25. Add AND subtract this value inside the second parenthesis:
Add AND subtract 25 in the second parenthesis.
Factor to get
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Distribute
Multiply
Combine like terms.
Add 625 to both sides.
Combine like terms.
Divide both sides by 400 (to make the right side equal to 1)
Break up the fraction.
Reduce
Rewrite 16 as . Rewrite 25 as
Rewrite as
Now the equation is in the form (which is the standard form of an ellipse) where , , and
So the value of is
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