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# Find the equation of the ellipse if the foci are $(\pm 3 , 0 ),$ and the length of the latus rectum is $\large\frac{32}{5}$

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}$$=1 • http://clay6.com/mpaimg/3_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/3_2_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)
Step 1:
The foci are at $(\pm,0)$.The length of the $LR=\large\frac{32}{5}$.
The centre is the midpoint of the line joining the foci.The centre is $(0,0)$.
$CF=ae=3$
The foci lie on the major axes,i.e.the $x$-axis.
Step 2:
Length of $LR=\large\frac{2b^2}{a}=\frac{32}{5}$
$b^2=\large\frac{16a}{5}$
Now $b^2=a^2(1-e^2)$
Therefore $a^2(1-e^2)=\large\frac{16a}{5}$
$a^2-a^2e^2=\large\frac{16a}{5}$
Now $ae=3\Rightarrow a^2-9=\large\frac{16a}{5}$
Step 3:
$5a^2-16a-45=0$
$5a^2-25a+9a-45=0$
$5a(a-5)+9(a-5)=0$
$(5a+9)(a-5)=0$
Since $'a'$ cannot be negative,$a=5$
Therefore $b^2=\large\frac{80}{5}$$=16 The equation is \large\frac{x^2}{25}+\frac{y^2}{16}$$=1$