Question

Find an equation of the ellipse that has center (1.-4), a major axis of length 12, and endpoint of minor axis (4.-4)

Answer 1

Solution:

Given:

Ellipse with the following properties;

$\begin{gathered} \text{centre}=(1,-4) \\ \text{length of major axis= 12} \\ \text{End point of minor axis = (4,-4)} \end{gathered}$

The equation of an ellipse is given by;

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where,

$\begin{gathered} (h,k)\text{ is the centre} \\ (h,k)=(1,-4) \\ h=1 \\ k=-4 \end{gathered}$

Substituting these values into the equation of an ellipse,

$\begin{gathered} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \\ \frac{(x-1)^2}{a^2}+\frac{(y-(-4))^2}{b^2}=1 \\ \frac{(x-1)^2}{a^2}+\frac{(y+4)^2}{b^2}=1 \end{gathered}$

The length of the semi-minor axis (a) is the difference between the x-values of the centre and the endpoint of the minor axis.

Hence,

$\begin{gathered} a=4-1 \\ a=3 \\ a^2=3^2 \\ a^2=9 \end{gathered}$

To get the length of the semi-major axis (b),

$\begin{gathered} \text{length of the major axis=2b} \\ 12=2b \\ b=\frac{12}{2} \\ b=6 \\ b^2=6^2 \\ b^2=36 \end{gathered}$

Substituting these into the equation of the ellipse,

$\begin{gathered} \frac{(x-1)^2}{a^2}+\frac{(y+4)^2}{b^2}=1 \\ \frac{(x-1)^2}{9^{}}+\frac{(y+4)^2}{36^{}}=1 \end{gathered}$

Therefore, the equation of the ellipse is;

$\frac{(x-1)^2}{9^{}}+\frac{(y+4)^2}{36^{}}=1$