# Find the equation for the ellipse that satisfies the given conditions : Ends of major axis $(0, \pm \sqrt 5 )$, ends of minor axis $(\pm 1, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{5}+\large\frac{y^2}{1}=1 & \quad (B)\; \large\frac{x^2}{1}+\large\frac{y^2}{5}=1 \\ (C)\;\large\frac{x^2}{1}-\large\frac{y^2}{5}=1 & \quad (D)\;\large\frac{x^2}{5}-\large\frac{y^2}{1}=1 \end {array}$

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 :
It is given that the ends of major axis is $(0, \pm \sqrt 5)$ and ends of minor axis is $( \pm 1, 0)$
$\therefore$ It is clear that the major axis is along the y - axis.
Hence the equation should be of the form $\large\frac{x^2}{b^2}$$+ \large\frac{Y^2}{a^2}$$=1$
hence $a^2= (\sqrt 5 )^2$ and $b^2 = (1)^2$
$\therefore$ Equation of the required ellipse is
$\large\frac{x^2}{1}$$+ \large\frac{y^2}{5}$$=1$