Question

cises 12.5 slete the following: Find the intercepts and domain, and perform the symmetry test on each of the following ellipses. x² + 4y² = 36

Answer 1

Given the equation of the ellipse :

$x^2+4y^2=36$

Divide the equation by 36:

$\begin{gathered} \frac{x^2}{36}+\frac{4y^{}2}{36}=\frac{36}{36} \\ \\ \frac{x^2}{36}+\frac{y^2}{9}=1 \\ \\ \frac{x^2}{6^2}+\frac{y^2}{3^2}=1 \end{gathered}$

The last equation is similar to :

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

As a > b , the major axis will parallel to x - axis

$\begin{gathered} a=6 \\ b=3 \end{gathered}$

The vertices will represents x- intercepts =

$(-6,0),(6,0)$

The y - intercepts will be :

$(0,-3),(0,3)$

The axis of symmetry will be the lines :

$\begin{gathered} x=0 \\ y=0 \end{gathered}$