# Find the equation of the ellipse if the foci are $(\pm 3 , 0 )$ and the vertices are $(\pm 5 , 0)$

Toolbox:
• Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
• $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{y^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/Toolbox1.png
• Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
• End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae) and (\pm \large\frac{b^2}{a}$$,-ae)$
• Directrices $y=\pm \large\frac{a}{e}$.
• The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
• General form of standard ellipses with centre $C$ with major axis $2a$,minor axis $2b$ and axes parallel to the coordinate axes.
• $\large\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$$=1 • In this case major axis y=k and minor axis x=h. • \large\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}$$=1$
• In this case major axis $x=h$ and minor axis $y=k$.
Step 1:
The foci are $F(3,0),F'(-3,0)$.
Therefore the $x$-axis($y$=0) is the major axis and the midpoint of $FF'$ (i.e) the origin (0,0) is the centre.
The vertices are $A(5,0),A'(-5,0)$
$OA=a=5$
$OF=ae=3$
Substituting $a=5$ we get
$e=\large\frac{3}{5}$
Step 2:
$b=a\sqrt{1-c^2}$
$\;\;=5\sqrt{1-\large\frac{9}{25}}$
$\;\;=4$
Step 3:
The equation is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 \large\frac{x^2}{5^2}+\frac{y^2}{4^2}$$=1$
$\large\frac{x^2}{25}+\frac{y^2}{16}$$=1$
This is the required equation.
edited Jun 13, 2013