If `tan alpha tan beta=-(a^(2))/(b^(2)`, then the chord joining two points alpha and beta on the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`, will subtend a right angle at

Options

A. focus
B. centre
C. end of the major axis
D. end of the minor axis

Answer

D

Explanation

Let `P (a cos theta_(1), b sin theta_(1))` and `Q (a cos theta_(2), b sin theta_(2))` be two points on the ellipse.
Then, `m_(1) =` slope of `OP = b/a tan theta_(1)`
and, `m_(2) = ` slope of `OQ = b/a tan theta_(2)`
`therefore m_(1)m_(2) = b^(2)/a^(2) tan theta_(1) tan theta_(2)`
`rArr m_(1)m_(2) = b^(2)/a^(2) xx (-a^(2))/(b^(2)) " " [because tan theta_(1) tan theta_(2) = -a^(2)//b^(2)` (given)]
`rArr m_(1)m_(2) = -1`
`therefore PAQ = pi//2`.
Hence, PQ makes a right angle at the centre of the allipse.