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# Find the equation of the ellipse if the vertices are $(\pm 4 , 0)$and $e=\large\frac{\sqrt{3}}{2}$

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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}$$=1 • http://clay6.com/mpaimg/4_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/4_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 verices are $(\pm 4,0)$ and $e=\large\frac{\sqrt 3}{2}$
Since the centre is the mid point of the line joining the vertices,the centre is at $(0,0)$.
Also $a=CA=4$
$b=a\sqrt{1-e^2}$
$\Rightarrow 4\sqrt{1-\large\frac{3}{4}}$
$\Rightarrow \large\frac{4}{2}$$=2 Step 2: Vertices lie on the major axis is the x-axis. Therefore the equation is \large\frac{x^2}{16}+\frac{y^2}{4}$$=1$
answered Jun 17, 2013