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ellipse x^2 + 2y^2 -6x+4y+7=0: find vertices, foci, graph
- Thread starter summergrl
- Start date
The ellipse equation you derived is incorrect. Sorry.
It should be:
\(\displaystyle \L\\\frac{(x-3)^{2}}{4}+\frac{(y+1)^{2}}{2}=1\)
It's center is at (3,-1)
Since x has the largest denominator, the major axis lies along the x-axis.
\(\displaystyle a=2, \;\ b=\sqrt{2}\)
The foci lie c units on either side of the center.
You have a and b. Use Pythagoras to find c.
It should be:
\(\displaystyle \L\\\frac{(x-3)^{2}}{4}+\frac{(y+1)^{2}}{2}=1\)
It's center is at (3,-1)
Since x has the largest denominator, the major axis lies along the x-axis.
\(\displaystyle a=2, \;\ b=\sqrt{2}\)
The foci lie c units on either side of the center.
You have a and b. Use Pythagoras to find c.
By completing the square with the expression you posted.
\(\displaystyle \L\\x^{2}+2y^{2}-6x+4y=-7\)
Group and complete square:
\(\displaystyle \L\\(x^{2}-6x+ \;\ )+(2y^{2}+4y+ \;\ )=-7\)
\(\displaystyle \L\\x^{2}-6x+9+2(y^{2}+2y+1)=-7+9+2\)
Factor:
\(\displaystyle \L\\(x-3)^{2}+2(y+1)^{2}=4\)
Divide by 4:
\(\displaystyle \L\\\frac{(x-3)^{2}}{4}+\frac{(y-1)^{2}}{2}=1\)
\(\displaystyle \L\\x^{2}+2y^{2}-6x+4y=-7\)
Group and complete square:
\(\displaystyle \L\\(x^{2}-6x+ \;\ )+(2y^{2}+4y+ \;\ )=-7\)
\(\displaystyle \L\\x^{2}-6x+9+2(y^{2}+2y+1)=-7+9+2\)
Factor:
\(\displaystyle \L\\(x-3)^{2}+2(y+1)^{2}=4\)
Divide by 4:
\(\displaystyle \L\\\frac{(x-3)^{2}}{4}+\frac{(y-1)^{2}}{2}=1\)