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Find the distance between foci of the ellipse \(\rm {x^2\over100}+{y^2\over64} = 1\).1. 22. 33. 44. 12 |
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Answer» Correct Answer - Option 4 : 12 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 ellipse \(\rm {x^2\over100}+{y^2\over64} = 1\) The eccentricity (e) ⇒ e = \(\rm \sqrt{100-64}\over 10\) ⇒ e = \(\rm \sqrt{36}\over 10\) ⇒ e = 0.6 Now distance between foci = 2ae = 2 × 10 × 0.6 ∴ Distance between foci = 12 |
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