__Category__: Analytic Geometry, Algebra"Published in Newark, California, USA"

Find the center, foci, and semi-axes for the ellipse and sketch the graph

__Solution__:

From the given equation above, the curve is ellipse because both x

^{2}and y

^{2}are positive but their coefficients are different. The general equation of an ellipse must be simplified in standard form in order to get its center, foci, and semi-axes as follows

Group the equation according to their variables and transpose the coefficient to the right side of the equation

Complete the square of each group

Divide both sides of the equation by 6

The above equation is now simplified in standard form. Since the denominator at x group is greater than the denominator at y group, then the major axis is parallel to x-axis.

To solve for the coordinates of the center:

Equate x + 2 = 0 Equate y + 1 = 0

x = -2 y = -1

Therefore, the coordinates of the center is

**C(-2, -1)**.

Next, solve for the value of semi-major axis

Therefore, the coordinates of the ends of major axis are

**V**and

_{1}(-4.4495, -1)**V**.

_{2}(0.4495, -1)Next, solve for the value of semi-minor axis

Therefore, the coordinates of the ends of minor axis are

**V**and

_{3}(-2, 0.4142)**V**.

_{4}(-2, -2.4142)Next, solve for the distance of a focus to the center

Therefore, the coordinates of the foci are

**F**and

_{1}(-4, -1)**F**.

_{2}(0, -1)Next, solve for the length of a semi-latus rectum

Add and subtract this value to the y values of foci in order to get the coordinates of the ends of latera recta. Therefore, the coordinates of the ends of latera recta are

**L**,

_{1}(-4, 0.1835)**L**,

_{2}(-4, -1.8165)**L**, and

_{3}(0, 0.1835)**L**.

_{4}(0, -1.8165)Since we know the coordinates of the center, ends of major axis, ends of minor axis, foci, and ends of latera recta, then we can sketch the graph of an ellipse as follows

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