# Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 6, 0), foci (\pm 4, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{36}-\large\frac{y^2}{20}=1 & \quad (B)\;\large\frac{x^2}{20}-\large\frac{y^2}{36}=1 \\ (C)\;\large\frac{x^2}{36}+\large\frac{y^2}{20}=1 & \quad (D)\;\large\frac{x^2}{20}+\large\frac{y^2}{36}=1 \end {array}$

Toolbox:
• If the major axis is along x - axis then the equation of ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
• If the major axis is along y - axis then the equation of ellipse is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
• $c = \sqrt{a^2+b^2}$ , where $c$ is the foci of the ellipse.
Step 1 :
The given vertices are $(\pm 6, 0)$ and foci $(\pm 4, 0)$
It is clear thta the vertices are on the x axis.
Hence the major axis is along x - axis.
$\therefore$ The equation of the ellipse should be of the form
$\large\frac{x^2}{a^2}$$+ \large\frac{y^2}{b^2}$$=1$
$b^2=a^2-c^2$
$\Rightarrow b^2 = 36-16$
$\therefore b = \sqrt{20}$
Hence the equation of ellipse is
$\large\frac{x^2}{36}$$+\large\frac{y^2}{20}$$=1$