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The tangent at the point alpha on the el...

The tangent at the point `alpha` on the ellipse `x^2/a^2+y^2/b^2=1` meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity 'e' of the ellipse is given by the equation (A) `e^2(1+cos^2alpha)=1` (B) `e^2(cosec^2alpha-1)=1` (C) `e^2(1+sin^2alpha)=1 (D) `e^2(1+tan^2alpha)=1`

Text Solution

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We can create the diagram as per thegiven details. Please refer to the diagram in the video.
Now, from the figure, we have
`AF_1 = a-aecostheta and AF_2 = a+aecostheta`(Here, e is eccentricity of ellipse)
`F_1F_2 = 2c`
Also,`F_1F_2^2 = AF_1^2 + AF_2^2`
`=>4c^2 = a^2+a^2e^2cos^2theta-2a^2ecostheta+a^2+a^2e^2cos^2theta+2a^2ecostheta`
Putting, `c^2 = a^2-b^2`
`=>4a^2-4b^2 = a^2+a^2e^2cos^2theta+a^2+a^2e^2cos^2theta`
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