SOLUTION: identify the conic section. if it is a parabola, give the vertex. if it is a circle, gibe the center and radius. if it is an ellipse or a hyperbola, give the center and foci.
4x2
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Question 384600: identify the conic section. if it is a parabola, give the vertex. if it is a circle, gibe the center and radius. if it is an ellipse or a hyperbola, give the center and foci.
4x2 + 7y2 - 40x + 42y +135 =0
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
There is an 
term AND a 
term -- not a parabola.
The signs on the and terms are the same -- not a hyperbola.
The coefficients on the and terms are not equal -- not a circle.
Therefore, it is an ellipse.
Complete the squares on each of the variables.
First move the constant term to the RHS and group like variables:
Complete the square on x:
Factor the coefficient on the term out of the and 
terms:
\ +\ 7y^2\ +\ 42y\ =\ -135)
Divide the coefficient on by 2, square the result, and add that result to the LHS inside the parentheses, then add the lead coefficient times the result of squaring half of the coefficient to the RHS. -10 divided by 2 is -5. -5 squared is 25. Add 25 inside the parentheses on the left and 100 to the right.
\ +\ 7y^2\ +\ 42y\ =\ -135\ +\ 100)
Repeat the same process for the y terms:
\ +\ 7\left(y^2\ +\ 6y\ +9\right)\ =\ -135\ +\ 100\ +\ 63)
Factor the two perfect square trinomials in the LHS and do the arithmetic in the RHS:
^2\ +\ 7\left(y\ +\ 3\right)\ =\ 28)
Divide both sides by the RHS constant.
^2}{7}\ +\ \frac{\left(y\ +\ 3\right)}{4}\ =\ 1)
Since there was no 
term in the original equation, the axes of the ellipse are parallel to the coordinate axes. Since the denominator in the term is larger than the denominator in the 
term, the major axis is parallel to the axis.
^2}{a^2}\ +\ \frac{\left(y\ -\ k\right)}{b^2}\ =\ 1)
Is an equation of an ellipse with center at:
)
Semi-major axis 
, semi-minor axis 
Vertices:
)
Foci:
)
where
And while we are at it, the endpoints of the minor axis segement are:
)
For your problem, you will find it convenient to rewrite the standard form equation thusly:
^2}{(\sqrt{7})^2}\ +\ \frac{\left(y\ -\ (-3)\right)}{2^2}\ =\ 1)
You can see that the center is at:
)
Then, to fix the foci, calculate 
So the foci are at:
)
John
My calculator said it, I believe it, that settles it
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