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# Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram: $16x^{2}+25y^{2}=400$

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}$$=1 • http://clay6.com/mpaimg/13_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/13_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:
$16x^2+25y^2=400$
The above equation is divided by 400 we get
$\large\frac{x^2}{25}+\frac{y^2}{16}$$=1 a^2=25,b^2=16 \Rightarrow a=5,b=4 The major axis is the x-axis. Step 2: Eccentricity :e=\sqrt{1-\large\frac{b^2}{a^2}} \Rightarrow \sqrt{1-\large\frac{16}{25}} \Rightarrow \sqrt{\large\frac{9}{25}} \Rightarrow \large\frac{3}{5} Step 3: Centre : The origin (0,0). Step 4: Foci (\pm ae,0). ae=5\times \large\frac{3}{5}$$=3$
The foci are $(\pm 3,0)$
Step 5:
Vertices : $(\pm a,0)$
$a=5$
Therefore the vertices are $(\pm 5,0)$
edited Jun 21, 2013