How do you graph the ellipse x^2/169+y^2/25=1 and find the center, the major and minor axis, vertices, foci and eccentricity?

Aug 6, 2017

Explanation:

This is the equation of an ellipse of the form ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$, whose center is $\left(h , k\right)$, major axis is $2 a$ and minor axis is $2 b$. Vertices on major axis are $\left(h \pm a , k\right)$ and along minor axis are $\left(h , k \pm b\right)$. Eccentricity is given by $e = \sqrt{1 - {b}^{2} / {a}^{2}}$ and focii are $\left(h \pm a e , k\right)$.

As ${x}^{2} / 169 + {y}^{2} / 25 = 1$ can be written as

${\left(x - 0\right)}^{2} / {13}^{2} + {\left(y - 0\right)}^{2} / {5}^{2} = 1$

Hence, this is an equation of an ellipse, whose center is $\left(0 , 0\right)$, major axis is $13 \times 2 = 26$ and minor axis is $5 \times 2 = 10$.

Vertices are $\left(- 13 , 0\right) , \left(13 , 0\right) , \left(5 , 0\right)$ and $\left(- 5 , 0\right)$.

Eccentricity is $e = \sqrt{1 - {5}^{2} / {13}^{2}} = \sqrt{{12}^{2} / {13}^{2}} = \frac{12}{13}$

and fociie are $\left(\pm \left(13 \times \frac{12}{13}\right) , 0\right)$ i.e. $\left(- 12 , 0\right)$ and $\left(12 , 0\right)$.

We can mark the four vertices, if so desired a few more points by using the equation ${x}^{2} / 169 + {y}^{2} / 25 = 1$. For example, if $x = \pm 3$. $y = \pm 5 \sqrt{1 - \frac{9}{169}} = \pm 20 \sqrt{\frac{10}{169}} = \pm 4.865$ gives four more points $\left(3 , 4.865\right) , \left(- 3 , 4.865\right) , \left(3 , - 4.865\right)$ and $\left(- 3 , - 4.865\right)$, joining which along with vertices will give us the desired ellipse.

The ellipse appears as shown below:

graph{(x^2/169+y^2/25-1)((x+13)^2+y^2-0.04)((x-13)^2+y^2-0.04)(x^2+(y+5)^2-0.04)(x^2+(y-5)^2-0.04)((x+12)^2+y^2-0.04)((x-12)^2+y^2-0.04)=0 [-20, 20, -10, 10]}