Ellipse: Standard Equation
DEFINITION OF AN ELLIPSE:
The set of all points (x, y) in a plane the sum of whose distances from two fixed points, called foci, is constant.
Figure 1 shows a picture of an ellipse:
Thus for all (x, y), d_{1} + d_{2} = constant. When talking about an ellipse, the following terms are used:
- The foci are two fixed points equidistant from the center of the ellipse.
- The vertices are the points on the ellipse that fall on the line containing the foci.
- The line segment or chord joining the vertices is the major axis.
- The midpoint of the major axis is the center.
- The axis perpendicular to the major axis is the minor axis.
Figure 1 shows an ellipse with a horizontal major axis. However an ellipse could also have a major axis that is vertical as shown in Figure 2.
The standard equation of an ellipse is:
STANDARD EQUATION OF AN ELLIPSE:
➢ Center coordinates (h, k)
➢ Major axis 2a
➢ Major axis 2b
➢ 0 < b < a
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ major axis is horizontal
$\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1$ major axis is vertical
The foci lie on the major axis, c units from the center, with c^{2} = a^{2} - b^{2}.
Let's use these equations in some examples:
Step 1: Determine the following: ➢ The coordinates of the center (h, k). ➢ The distance of half the minor axis (b). ➢ The length of half the major axis (a). ➢ The orientation of the major axis. |
Center: Since the foci are equidistant from the center of the ellipse the center can be determine by finding the midpoint of the foci. $\left(h,k\right)=\left(\frac{2+\left(-4\right)}{2},\frac{1+1}{2}\right)=\left(-\frac{2}{2},\frac{2}{2}\right)={\left(}-1,1{\right)}$ Length of b: The minor axis is given as 10, which is equal to 2b. $2b=10\to {b}={5}$ Length of a: To find a the equation c^{2} = a^{2} + b^{2} can be used but the value of c must be determined. Since c is the distance from the foci to the center, take either foci and determine the distance to the center. Then solve for a. Foci (2, 1): $c=\left|2-\left(-1\right)\right|=\left|3\right|=3$ Foci (-4, 1): $c=\left|-4-\left(-1\right)\right|=\left|-3\right|=3$ c = 3 ${c}^{2}={a}^{2}-{b}^{2}\to a=\sqrt{{c}^{2}+{b}^{2}}$ $a=\sqrt{{{3}}^{2}+{{5}}^{2}}\to {a}{=}\sqrt{34}$ Orientation of major axis: Since the two foci fall on the horizontal line y = 1, the major axis is horizontal. |
Step 2: Substitute the values for h, k, a and b into the equation for an ellipse with a horizontal major axis. |
Horizontal major axis equation: $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ Substitute values: $\frac{{\left[x-\left(-1\right)\right]}^{2}}{{\sqrt{34}}^{2}}+\frac{{\left(y-1\right)}^{2}}{{{5}}^{2}}=1$ Simplify: $\frac{{\left(x+1\right)}^{2}}{{\sqrt{34}}^{2}}+\frac{{\left(y-1\right)}^{2}}{{5}^{2}}=1$ |
Step 1: Group the x- and y-terms on the left-hand side of the equation. |
x^{2} + 3y^{2} - 4x - 18y + 4 = 0 $({x}^{2}-4x)+(3{y}^{2}-18y)+4=0$ |
Step 2: Move the constant term to the right-hand side. |
$({x}^{2}-4x)+(3{y}^{2}-18y)=-4$ |
Step 3: Complete the square for the x- and y-groups. |
$({x}^{2}-4x)+(3{y}^{2}-18y)=-4$ Complete the square for the x-group (x^{2} - 4x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{-4}{2}=-2;{\left(-2\right)}^{2}=4$ Add the result to the x-group. (x^{2} - 4x + 4) Complete the square for the y-group (3y^{2} - 18y) Factor out a 3 so they y^{2}-coefficient is 1 3(y^{2} - 6y) Take the coefficient of the y-term, divide by 2 and square the result. $\frac{-6}{2}=-3;{\left(-3\right)}^{2}=9$ Add the result to the y-group. 3(y^{2} - 6y + 9) Final result. $({x}^{2}-4x+4)+3({y}^{2}-6y+9)=-4+?$ |
Step 4: Add whatever was added to the left-hand side to the right-hand side. |
$({x}^{2}-4x+4)+3({y}^{2}-6y+9)=-4+4+3\left(9\right)$ $({x}^{2}-4x+4)+3({y}^{2}-6y+9)=27$ |
Step 5: Write the x-group and y-group as perfect squares. |
${\left(x-2\right)}^{2}+3{\left(y-3\right)}^{2}=27$ |
Step 6: Divide both sides by the value on the right-hand side. |
$\frac{{\left(x-2\right)}^{2}}{27}+\frac{3{\left(y-3\right)}^{2}}{27}=1$ $\frac{{\left(x-2\right)}^{2}}{27}+\frac{{\left(y-3\right)}^{2}}{9}=1$ |
Related Links: Math algebra Ellipse: Eccentricity Hyperbola: Standard Equation Pre Calculus |
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