If e is the eccentricity of the ellipse x2a2+y2b2=1 a<b, then(a) b2 = – InterviewSolution

1.	If e is the eccentricity of the ellipse x2a2+y2b2=1 a<b, then(a) b2 = a2(1 – e2)(b) a2 = b2(1 – e2)(c) a2 = b2(e2 – 1)(d) b2 = a2(e2 – 1)
Answer» If e is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b),$ then (a) b² = a²(1 – e²) (b) a² = b²(1 – e²) (c) a² = b²(e² – 1) (d) b² = a²(e² – 1)

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