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Madhavi Gagrani
Asked: 2022-11-09T06:49:04+05:30 In:

What is the equation of the ellipse having foci (±2,0) and the eccentricity 1/2
1. \(\rm \frac{x^2}{6}-\frac{y^2}{16}=1\)
2. \(\rm \frac{x^2}{12}+\frac{y^2}{16}=1\)
3. \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)
4. \(\rm \frac{x^2}{4}+\frac{y^2}{144}=1\)

What is the equation of the ellipse having foci (±2,0) and the eccentricity 1/2
1. \(\rm \frac{x^2}{6}-\frac{y^2}{16}=1\)
2. \(\rm \frac{x^2}{12}+\frac{y^2}{16}=1\)
3. \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)
4. \(\rm \frac{x^2}{4}+\frac{y^2}{144}=1\)
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  1. Correct Answer – Option 3 : \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)

    Concept:

    Equation of ellipse: \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

    Foci: (±ae, 0)

    Eccentricity, e = \(\rm \sqrt{1-\frac{b^2}{a^2}}\)

     

    Calculation:

    Here,foci = (±2,0) = (±ae, 0) and the eccentricity, e = 1/2

    ae = 2 

    ⇒ a × 1/2 = 2 

    ⇒ a = 4

    ⇒ a2 = 16

    Now, e = \(\rm \sqrt{1-\frac{b^2}{a^2}}\)

    ⇒ b2= a2(1 – e2)

    = 16(1 – 1/4)

    = 16× (3/4)

    = 12

    ∴ Equation of ellipse = \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)

    Hence, option (3) is correct.

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