1.

What is the equation of the ellipse having foci (±2,0) and the eccentricity 1/21. \(\rm \frac{x^2}{6}-\frac{y^2}{16}=1\)2. \(\rm \frac{x^2}{12}+\frac{y^2}{16}=1\)3. \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)4. \(\rm \frac{x^2}{4}+\frac{y^2}{144}=1\)

Answer» 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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