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Mathematics - Ellipse Question with Solution | TestHub

MathematicsEllipseGeneralHard3 minQB

MathematicsHardmultiple choice

Passage / Comprehension

The ellipse $x^{2} / (lo g a)^{2} + y^{2} / (lo g b)^{2} = 1$ has eccentricity $1/ 2$ , and a related hyperbola is $x^{2} / (lo g_{b} a)^{2} - y^{2} = 1$ .

If its eccentricity is $e$ , then $e^{2}$ is greater than which values?

Options:(select one or more)

Answer:

B, C, D

Solution:

$(lo g b)^{2} = (lo g a)^{2} /2$ , so $(lo g_{b} a)^{2} = 2$ and $e^{2} = 1 + 1/2 = 3/2$ , which exceeds B, C and D.

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:General

⏱ 3mℹ️ Source: QB

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