The tangent at a point `P(acostheta,bsintheta)` of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` meets its auxiliary circle at two points, the chord joining which subtends a right angle at the center. Find the eccentricity of the ellipse.

Options

A. `(1 + sin^(2) theta)^(-1//2)`
B. `(1 + cos^(2) theta)^(-1//2)`
C. `(1 + sin^(2) theta)`
D. `(1 + cos^(2) theta)^(1//2)`

Answer

A

Explanation

The equation of the tangent at `P (a cos theta, b sin theta)` to the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` is
`x/a cos theta + y/b sin theta = 1`
The combined equation of the chords joining the points of intersection of (i) and the auxiliary circle `x^(2) + y^(2) = a^(2)` is
`x^(2) + y^(2) = a^(2) (x/a cos theta + y/b sin theta)^(2)`
The lines given by the equation are at right angles.
`therefore` Coeff. of `x^(2) +` Coeff. of `y^(2) 0`
`rArr 1 - a^(2)((cos^(2)theta)/(a^(2))) + 1 - a^(2)((sin^(2)theta)/(b^(2))) = 0`
`rArr sin^(2)theta + 1 - a^(2)/b^(2) sin^(2)theta = 0`
`rArr sin^(2)theta(1 - a^(2)/b^(2)) + 1 = 0`
`rArr - sin^(2)theta(a^(2) - b^(2)) + b^(2) = 0`
`rArr - a^(2)e^(2) sin^(2)theta + a^(2)(1 - e^(2)) = 0`
`rArr 1 = e^(2) (1 + sin^(2)theta) rArr e = (1 + sin^(2)theta)^(-1//2)`