A point on the ellipse `x^(2)/16 + y^(2)/9 = 1` at a distance equal to the mean of lengths of the semi - major and semi-minor axis from the centre, is

Options

A. `((2sqrt(91))/(7), (3sqrt(105))/(14))`
B. `((2sqrt(91))/(7), (-3sqrt(91))/(14))`
C. `((-2sqrt(105))/(7), (-3sqrt(91))/(14))`
D. `((-2sqrt(105))/(7), (sqrt(91))/(14))`

Answer

A

Explanation

Let `P (4 cos theta, 3 sin theta)` be a point on the given ellipse such that its distance from the centre (0, 0) of the ellipse is equal to the mean of the lengths of the semi-major and semi-minor axes i.e.
`OP = (4 + 3)/(2)`
`rArr Sqrt(16cos^(2)theta + 9 sin^(2)theta) = 7/2`
`rArr 7 cos^(2)theta + 9 = 49/4`
`rArr cos^(2)theta = 13/28`
`rArr cos theta = pm sqrt(13/28) and sin theta = pm sqrt(15/28)`
`rArr cos theta = pm (sqrt(91))/(14) and sin theta = pm (sqrt(105))/(14)`
Hence, the required points are given by
`P(pm (2sqrt(91))/(7), pm (3sqrt(105))/(14))`