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Find the equation of ellipse whose eccentricity is \(\rm 1\over2\), length of latus rectum is 4 and the center is (0, 0).


1. \(\rm 9x^2+12y^2 = 64\)
2. \(\rm 9x^2+12y^2 = 16\)
3. \(\rm 12x^2+9y^2 = 64\)
4. \(\rm 12x^2+9y^2 = 16\)

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Correct Answer - Option 1 : \(\rm 9x^2+12y^2 = 64\)

Concept:

The standard equation of an ellipse:

\(\rm {x^2\over a^2}+{y^2\over b^2} = 1\)

Where 2a and 2b are the length of the major axis and minor axis respectively and center (0, 0)

The eccentricity = \(\rm \sqrt{(a^2-b^2)}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2-b^2}\)

Calculation:

Given Eccentricity (e) = \(\rm 1\over2\)

⇒ \(\rm {\sqrt{a^2-b^2}\over a} = {1\over2}\) 

⇒ \(\rm {a^2-b^2} = {a^2\over4}\)

⇒ \(\boldsymbol{\rm b^2 = {3a^2\over4}}\)             ...(i)

Also Latus rectum length = 4

⇒ \(\rm {2b^2\over a} = 4\)

⇒ \(\boldsymbol{\rm b^2= 2a}\)              ...(ii)

Comparing (i) and (ii)

⇒ \(\rm {3a^2\over4} = 2a\)

a = \(\rm 8\over3\)

\(\rm b^2= 2a = {16\over3}\)

a2 = \(\rm 64\over9\)

The equation of ellipse 

\(\rm {x^2\over a^2}+{y^2\over b^2} = 1\)

⇒ \(\boldsymbol{\rm 9x^2+12y^2 = 64}\)

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