How do you graph  (x - 2)^2 + 3(y - 2)^2 = 25?

Jul 22, 2017

Explanation:

${\left(x - 2\right)}^{2} + 3 {\left(y - 2\right)}^{2} = 25$ is the equation of an ellipse, as can be seen from the following

${\left(x - 2\right)}^{2} + 3 {\left(y - 2\right)}^{2} = 25$

$\implies {\left(x - 2\right)}^{2} / 25 + 3 {\left(y - 2\right)}^{2} / 25 = 1$

or ${\left(x - 2\right)}^{2} / {5}^{2} + {\left(y - 2\right)}^{2} / {\left(\frac{5}{\sqrt{3}}\right)}^{2} = 1$

an ellipse, whose center is $\left(2 , 2\right)$ and major axis is $2 \times 5 = 10$ parallel to $x$-axis and minor axis is $2 \times \frac{5}{\sqrt{3}} = \frac{10}{\sqrt{3}}$ parallel to $x$-axis.

End points of major axis are $\left(2 \pm 5 , 2\right)$ i.e. $\left(7 , 2\right)$ and $\left(- 3 , 2\right)$

End points of minor axis are $\left(2 , 2 \pm \frac{5}{\sqrt{3}}\right)$ i.e. $\left(2 , 2 - \frac{5}{\sqrt{3}}\right)$ and $\left(2 , 2 + \frac{5}{\sqrt{3}}\right)$

It appears as follows:

graph{((x-2)^2+3(y-2)^2-25)((x+3)^2+(y-2)^2-0.02)((x-7)^2+(y-2)^2-0.02)((x-2)^2+(y-2+5/sqrt3)^2-0.02)((x-2)^2+(y-2-5/sqrt3)^2-0.02)=0 [-8.5, 11.5, -3.2, 6.8]}