Help me solve this math
- Thread starter Babalola
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Dr.Peterson
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I would first complete the square, in order to put it in standard form and find the semiaxes.
Then I'd look up the formula for eccentricity.
(By the way, this probably should be called algebra, or maybe geometry, rather than arithmetic. And the word "ellipse" in the title might have attracted attention quicker.)
HallsofIvy
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The equation is \(\displaystyle 9x^2+25y^2- 18x+150y+ 9=0\).
That can be written as \(\displaystyle 9(x^2- 2x)+ 25(y^2+ 6y)= -9\).
Complete the squares: \(\displaystyle 9(x^2- 2x+ 1- 1)+ 25(y^2- 6y+ 9- 9)= -9\).
\(\displaystyle 9(x^2- 2x+ 1)- 9+ 25(y^2- 6y+ 9)- 225= -9\)
\(\displaystyle 9(x- 1)^2+ 25(y- 3)^2= -9+ 9+ 225= 225\)
\(\displaystyle \frac{(x- 1)^2}{25}+ \frac{(y-3)^2}{9}= 1\).
That is an ellipse with center at (1, 3), vertices at (6, 3), (-4, 3), (1, 6) and (1, 0).
The "eccentricity" of the ellipse \(\displaystyle \frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1\) is given by \(\displaystyle e= \sqrt{1- \frac{b^2}{a^2}}\).
That can be written as \(\displaystyle 9(x^2- 2x)+ 25(y^2+ 6y)= -9\).
Complete the squares: \(\displaystyle 9(x^2- 2x+ 1- 1)+ 25(y^2- 6y+ 9- 9)= -9\).
\(\displaystyle 9(x^2- 2x+ 1)- 9+ 25(y^2- 6y+ 9)- 225= -9\)
\(\displaystyle 9(x- 1)^2+ 25(y- 3)^2= -9+ 9+ 225= 225\)
\(\displaystyle \frac{(x- 1)^2}{25}+ \frac{(y-3)^2}{9}= 1\).
That is an ellipse with center at (1, 3), vertices at (6, 3), (-4, 3), (1, 6) and (1, 0).
The "eccentricity" of the ellipse \(\displaystyle \frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1\) is given by \(\displaystyle e= \sqrt{1- \frac{b^2}{a^2}}\).