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# Find the equation of the ellipse if the centre is $(3 , -4 ),$one of the foci is $(3+\sqrt{3},-4)$ and $e=\large\frac{\sqrt{3}}{2}$

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}$$=1 • http://clay6.com/mpaimg/2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/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)
• 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:
Centre $C(3,-4)$ and one focus is $F(3+\sqrt 3,-4)$ with $e=\large\frac{\sqrt 3}{2}$
Now $CF=ae=\sqrt 3$
Therefore $a\large\frac{\sqrt 3}{2}$$=\sqrt 3 \Rightarrow a=2 Step 2: b=a\sqrt{1-e^2} \;\;=2\sqrt{1-\large\frac{3}{4}} \;\;=1 Step 3: The major axis is along CF whose equation is y=-4 which is parallel to the x-axis Therefore the equation of the ellipse is \large\frac{(x-a)^2}{a^2}+\frac{(y-a)^2}{b^2}$$=1$
$\large\frac{(x-3)^2}{4}+$$(y+4)^2=1$
This is the required equation.
edited Jun 13, 2013