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Exercise 5.6 · Q19

Q.An ellipse has OBOB as semi minor axes, FF and F′F' its foci and the angle FBF′FBF' is a right angle. Then the eccentricity of the ellipse is

(1) 12\dfrac1{\sqrt2}
(2) 12\dfrac12
(3) 14\dfrac14
(4) 13\dfrac1{\sqrt3}
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The distance from an endpoint of the minor axis to either focus is always exactly aa (a direct consequence of a2=b2+c2a^2=b^2+c^2); a right angle between two sides of length aa then fixes the base FF′=2cFF'=2c via Pythagoras.

Step 1. Distance from B(0,b)B(0,b) to a focus F(c,0)F(c,0).

BF=b2+c2BF=\sqrt{b^2+c^2}. Since for an ellipse a2=b2+c2a^2=b^2+c^2, BF=aBF=a. By symmetry BF′=aBF'=a too.

Step 2. Apply Pythagoras to the right angle at BB (legs BF=BF′=aBF=BF'=a, hypotenuse FF′=2cFF'=2c). …

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