Equation of ellipse with foci ( pm 2,0) and eccentricity = frac{1}{2}

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10-2. Parabola, Ellipse, Hyperbola

easy

What is the equation of the ellipse with foci $( \pm 2,\;0)$ and eccentricity $ = \frac{1}{2}$

A

$3{x^2} + 4{y^2} = 48$

B

$4{x^2} + 3{y^2} = 48$

C

$3{x^2} + 4{y^2} = 0$

D

$4{x^2} + 3{y^2} = 0$

Solution

(a) Since, $ae = \pm \,2$ $ \Rightarrow \,a = \pm \,4$

$(\because \,\,e = 1/2)$

Now ${b^2} = {a^2}(1 – {e^2})$

$ \Rightarrow $${b^2} = 16(1 – 1/4)$

$ \Rightarrow $${b^2} = 12$

Hence ellipse is$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1$

$ \Rightarrow $$3{x^2} + 4{y^2} = 48$.

Standard 11

Mathematics

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