Graph the ellipse given by the equation
$\text{\hspace{0.17em}}49{x}^{2}+16{y}^{2}=784.\text{\hspace{0.17em}}$ Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
When an
ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point,
$\text{\hspace{0.17em}}\left(h,k\right),$ we use the standard forms
$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{}ab\text{\hspace{0.17em}}$ for horizontal ellipses and
$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{}ab\text{\hspace{0.17em}}$ for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
Given the standard form of an equation for an ellipse centered at
$\text{\hspace{0.17em}}\left(h,k\right),$ sketch the graph.
Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
If the equation is in the form
$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where
$\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then
the center is
$\text{\hspace{0.17em}}\left(h,k\right)$
the major axis is parallel to the
x -axis
the coordinates of the vertices are
$\text{\hspace{0.17em}}\left(h\pm a,k\right)$
the coordinates of the co-vertices are
$\text{\hspace{0.17em}}\left(h,k\pm b\right)$
the coordinates of the foci are
$\text{\hspace{0.17em}}\left(h\pm c,k\right)$
If the equation is in the form
$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{\hspace{0.17em}}$ where
$\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then
the center is
$\text{\hspace{0.17em}}\left(h,k\right)$
the major axis is parallel to the
y -axis
the coordinates of the vertices are
$\text{\hspace{0.17em}}\left(h,k\pm a\right)$
the coordinates of the co-vertices are
$\text{\hspace{0.17em}}\left(h\pm b,k\right)$
the coordinates of the foci are
$\text{\hspace{0.17em}}\left(h,k\pm c\right)$
Solve for
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ using the equation
$\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.$
Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
Graphing an ellipse centered at (
h ,
k )
Graph the ellipse given by the equation,
$\text{\hspace{0.17em}}\frac{{\left(x+2\right)}^{2}}{4}+\frac{{\left(y-5\right)}^{2}}{9}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci.
First, we determine the position of the major axis. Because
$\text{\hspace{0.17em}}9>4,$ the major axis is parallel to the
y -axis. Therefore, the equation is in the form
$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{\hspace{0.17em}}$ where
$\text{\hspace{0.17em}}{b}^{2}=4\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{a}^{2}=9.\text{\hspace{0.17em}}$ It follows that:
the center of the ellipse is
$\text{\hspace{0.17em}}\left(h,k\right)=\left(\mathrm{-2},\text{5}\right)$
the coordinates of the vertices are
$\text{\hspace{0.17em}}(h,k\pm a)=(-2,5\pm \sqrt{9})=(-2,5\pm 3),$ or
$\text{\hspace{0.17em}}\left(\mathrm{-2},\text{2}\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(\mathrm{-2},\text{8}\right)$
the coordinates of the co-vertices are
$\text{\hspace{0.17em}}(h\pm b,k)=(-2\pm \sqrt{4},5)=(-2\pm 2,5),$ or
$\text{\hspace{0.17em}}\left(\mathrm{-4},5\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(0,\text{5}\right)$
the coordinates of the foci are
$\text{\hspace{0.17em}}\left(h,k\pm c\right),\text{\hspace{0.17em}}$ where
$\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ Solving for
$\text{\hspace{0.17em}}c,$ we have:
Therefore, the coordinates of the foci are
$\text{\hspace{0.17em}}\left(\mathrm{-2},\text{5}-\sqrt{5}\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(\mathrm{-2},\text{5+}\sqrt{5}\right).$
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Graph the ellipse given by the equation
$\text{\hspace{0.17em}}\frac{{\left(x-4\right)}^{2}}{36}+\frac{{\left(y-2\right)}^{2}}{20}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci.
Center:
$\text{\hspace{0.17em}}\left(4,2\right);\text{\hspace{0.17em}}$ vertices:
$\text{\hspace{0.17em}}\left(-2,2\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(10,2\right);\text{\hspace{0.17em}}$ co-vertices:
$\text{\hspace{0.17em}}\left(4,2-2\sqrt{5}\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(4,2+2\sqrt{5}\right);\text{\hspace{0.17em}}$ foci:
$\text{\hspace{0.17em}}\left(0,2\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(8,2\right)$
Given the general form of an equation for an ellipse centered at (
h ,
k ), express the equation in standard form.
Recognize that an ellipse described by an equation in the form
$\text{\hspace{0.17em}}a{x}^{2}+b{y}^{2}+cx+dy+e=0\text{\hspace{0.17em}}$ is in general form.
Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
Factor out the coefficients of the
$\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{y}^{2}\text{\hspace{0.17em}}$ terms in preparation for completing the square.
Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant,
$\text{\hspace{0.17em}}{m}_{1}{\left(x-h\right)}^{2}+{m}_{2}{\left(y-k\right)}^{2}={m}_{3},$ where
$\text{\hspace{0.17em}}{m}_{1},{m}_{2},$ and
$\text{\hspace{0.17em}}{m}_{3}\text{\hspace{0.17em}}$ are constants.
Divide both sides of the equation by the constant term to express the equation in standard form.