Lesson
Notes
| \({\text{Ellipses}}\) |
| \(\text{Equation}\) |
\(\displaystyle\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)
\(a\gt b\) |
\(\displaystyle\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\)
\(a\gt b\) |
| \(\text{Shape}\) |
 |
 |
| \(\text{Foci}\) |
\((h \pm c,k)\) |
\((h ,k \pm c)\) |
| \(\text{Vertices}\) |
\((h \pm a,k)\) |
\((h ,k \pm a)\) |
| \(\text{Center}\) |
\((h,k)\) |
| \(\text{Focal Length Equation}\) |
\(a^2-b^2=c^2\) |
| \(\text{Length of Major Axis}\) |
\(2a\) |
| \(\text{Length of Minor Axis}\) |
\(2b\) |
| \(\text{Eccentricity}\) |
\(\displaystyle\frac{c}{a}\) |
Questions
Graph the following ellipses.
\(\textbf{1)}\) \(\displaystyle\frac{x^2}{4}+\displaystyle\frac{y^2}{25}=1\)
This is the same as \(\displaystyle\left(\frac{x}{2}\right)^2+\displaystyle\left(\frac{y}{5}\right)^2=1\)
The center is \((0,0)\), x-radius is \(2\), y-radius is \(5\)
\(\textbf{2)}\) \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\)
This is the same as \(\displaystyle\left(\frac{x-3}{4}\right)^2+\displaystyle\left(\frac{y+2}{2}\right)^2=1\)
The center is \((3,-2)\), x-radius is \(4\) y-radius is \(2\)
\(\textbf{3)}\) Find the equation of an ellipse with vertices at \((3,9)\) and \((3,-1)\) and foci at \((3,8)\) and \((3,0)\).
The equation is \(\displaystyle\frac{(x-3)^2}{9}+\displaystyle\frac{(y-4)^2}{25}=1\)
For problems 4-7, use \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\)
\(\textbf{4)}\) Find the center of \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\).
\(\text{Hint: The center is } (h,k) \text{ in }\displaystyle\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)
The center is \((3,-2)\)
\(\textbf{5)}\) Find the vertices of \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\).
\(\text{Hint: The vertices are } (h+a,k) \text{ and } (h-a,k) \text{ in }\displaystyle\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)
The vertices are \((-1,-2)\) and \((7,-2)\)
\(\textbf{6)}\) Find the Foci of \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\).
The foci are \((3-2\sqrt{3},-2)\) and \((3+2\sqrt{3},-2)\)
\(\textbf{7)}\) Find the length of the minor axis of \(\displaystyle\frac{(x-3)^2}{16}+\displaystyle\frac{(y+2)^2}{4}=1\).
\(\text{Hint: The length of the minor axis is } 2b \text{ in }\displaystyle\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)
The length of minor axis is \(4\)
\(\textbf{8)}\) Use completing the square to rewrite the equation in standard form. \(9x^2+4y^2-8y=32\)
The equation is \(\displaystyle\frac{x^2}{4}+\displaystyle\frac{(y-1)^2}{9}=1\)
\(\,\,\,\,\,\,9x^2+4y^2-8y=32\)
\(\,\,\,\,\,\,9\left(x\right)^2+4\left(y^2-2y+\text{___}\right)=32+4\left(\text{___}\right)\)
\(\,\,\,\,\,\,9\left(x\right)^2+4\left(y^2-2y+\underline{ 1 }\right)=32+4\left(\underline{ 1 }\right)\)
\(\,\,\,\,\,\,9\left(x\right)^2+4\left(y-1\right)^2=36\)
\(\,\,\,\,\,\,\displaystyle\frac{9\left(x\right)^2}{36}+\frac{4\left(y-1\right)^2}{36}=\frac{36}{36}\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2}{4}+\frac{\left(y-1\right)^2}{9}=1\)
\(\textbf{9)}\) Write the equation of the following ellipse in standard form.

The equation is \(\displaystyle\frac{(x-2)^2}{16}+\displaystyle\frac{(y-3)^2}{4}=1\)
See Related Pages\(\)
