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Question

Question: What is the equation of the ellipse with foci \(( \pm 2,0)\) and eccentricity \(\frac{1}{2}\)...

What is the equation of the ellipse with foci (±2,0)( \pm 2,0) and eccentricity 12\frac{1}{2}

A

3x2+4y2=483x^{2} + 4y^{2} = 48

B

4x2+3y2=484x^{2} + 3y^{2} = 48

C

3x2+4y2=03x^{2} + 4y^{2} = 0

D

4x2+3y2=04x^{2} + 3y^{2} = 0

Answer

3x2+4y2=483x^{2} + 4y^{2} = 48

Explanation

Solution

Here ae=±2,e=12,a=±4ae = \pm 2,\because e = \frac{1}{2},\therefore a = \pm 4

Form b2=a2(1e2)b^{2} = a^{2}(1 - e^{2})b2=16(114)b^{2} = 16\left( 1 - \frac{1}{4} \right)b2=12b^{2} = 12

Hence, the equation of ellipse is x216+y212=1\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

or 3x2+4y2=483x^{2} + 4y^{2} = 48