Question

How do you use the definition of an ellipse and the distance formula to find an equation of the ellipse whose minor axis has length 12 and its focal points are at $\left( 1,2 \right)$ and $\left( 4,-2 \right)$?

Answer 1

We equate the given equation of elliptic curve with the general equation of $\dfrac{{{\left( x-\alpha \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-\beta \right)}^{2}}}{{{b}^{2}}}=1$. We find the length of the minor axis as $2b$ units and the distance between the two foci are $2ae=2\sqrt{{{a}^{2}}-{{b}^{2}}}$ units. We use those details to find the equation and plot on the graph.

Complete step-by-step answer:
The general equation of ellipse is $\dfrac{{{\left( x-\alpha \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-\beta \right)}^{2}}}{{{b}^{2}}}=1$. We also assume that ${{b}^{2}}<{{a}^{2}}$.
For the general equation $\left( \alpha ,\beta \right)$ is the centre which is the midpoint of foci. The vertices are $\left( \alpha \pm a,\beta \right)$. The coordinates of the foci are $\left( \alpha \pm ae,\beta \right)$. Here $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ is the eccentricity.
The centre is $\left( \dfrac{1+4}{2},\dfrac{2-2}{2} \right)\equiv \left( \dfrac{5}{2},0 \right)$.
We also know that the length of the minor axis is $2b$ units and the distance between the two foci are $2ae=2\sqrt{{{a}^{2}}-{{b}^{2}}}$ units.
For our given problem, minor axis has length 12 and its focal points are $\left( 1,2 \right)$ and $\left( 4,-2 \right)$.
Therefore, $2b=12$ which gives $b=\dfrac{12}{2}=6$.
We also know that for two points $\left( x,y \right)$ and $\left( c,d \right)$, the distance between them is $\sqrt{{{\left( x-c \right)}^{2}}+{{\left( y-d \right)}^{2}}}$ units.
The distance between $\left( 1,2 \right)$ and $\left( 4,-2 \right)$ is $\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}}=\sqrt{9+16}=\sqrt{25}=5$ units.
Also, $2ae=2\sqrt{{{a}^{2}}-{{b}^{2}}}=5$. We put the value of $b$ and get
$\begin{aligned} & 2\sqrt{{{a}^{2}}-{{6}^{2}}}=5 \\\ & \Rightarrow {{a}^{2}}-{{6}^{2}}={{\left( \dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4} \\\ & \Rightarrow {{a}^{2}}={{6}^{2}}+\dfrac{25}{4}=\dfrac{169}{4} \\\ \end{aligned}$
Now we put all the values to get the equation

& \dfrac{{{\left( x-\dfrac{5}{2} \right)}^{2}}}{\dfrac{169}{4}}+\dfrac{{{\left( y-0 \right)}^{2}}}{{{6}^{2}}}=1 \\\ & \Rightarrow \dfrac{{{\left( 2x-5 \right)}^{2}}}{169}+\dfrac{{{y}^{2}}}{36}=1 \\\ \end{aligned}$$  **Note:** The major axis goes through the foci and the centre. The inequality of the values of $a$ and $b$ creates the major and minor axes. If they were equal, the curve would have changed to circle from the ellipse.