An ellipse has OB as semi minor axis, F and F' its foci and the angle FBF' is a right angle. Then the eccentricity of the ellipse is: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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An ellipse has OB as semi minor axis, F and F^{'} its foci and the angle ∠ FB F^{'} is a right angle. Then the eccentricity of the ellipse is:

Solution & Step-by-step Explanation

Let the ellipse be \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 . B = (0, b), F = (a e, 0), F^{'} = (- a e, 0) .Since ∠ FB F^{'} = 9 0^{\circ} and △ FB F^{'} is symmetric about the y-axis, ∠ FBO = 4 5^{\circ} .In △ FBO (right-angled at O): tan (4 5^{\circ}) = \frac{OF}{OB} = \frac{a e}{b} 1 = \frac{a e}{b} ⟹ b = a e We know b^{2} = a^{2} (1 - e^{2}) : (a e)^{2} = a^{2} (1 - e^{2}) a^{2} e^{2} = a^{2} - a^{2} e^{2} ⟹ 2 a^{2} e^{2} = a^{2} 2 e^{2} = 1 ⟹ e^{2} = 1/2 ⟹ e = 1/ 2 .

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An ellipse has OB as semi minor axis, F and F^{'} its foci and the angle ∠ FB F^{'} is a right angle. Then the eccentricity of the ellipse is:

1/ 2

1/2

1/4

1/ 3

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