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The equation of the ellipse having foci (2, 0), ( -2, 0) and minor axis of length 8 units is:1. \(\rm \frac{x^2}{16}+\frac{y^2}{20}=1\)2. \(\rm \frac{x^2}{20}+\frac{y^2}{16}=1\)3. \(\rm \frac{x^2}{2\sqrt5}+\frac{y^2}{4}=1\)4. \(\rm \frac{x^2}{4}+\frac{y^2}{2\sqrt5}=1\) |
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Answer» Correct Answer - Option 2 : \(\rm \frac{x^2}{20}+\frac{y^2}{16}=1\) Concept: The distance between the centre and the focus of an ellipse is c = ae The equation of an ellipse with the length of the major axis 2a and the minor axis 2b is given by: \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
Calculation: Length of the minor axis = 2b = 8. ⇒ b = 4. Also, c = distance between the centre and the focus = ae = 2. c2 = a2e2 = a2 - b2 ∴ 22 = a2 - 42 ⇒ a2 = 20 Equation of the ellipse = \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). ⇒ \(\rm \frac{x^2}{20}+\frac{y^2}{16}=1\). |
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