Tangents are drawn to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1,(a > b),` and the circle `x^2+y^2=a^2` at the points where a common ordinate cuts them (on the same side of the x-axis). Then the greatest acute angle between these tangents is given by `tan^(-1)((a-b)/(2sqrt(a b)))` (b) `tan^(-1)((a+b)/(2sqrt(a b)))` `tan^(-1)((2a b)/(sqrt(a-b)))` (d) `tan^(-1)((2a b)/(sqrt(a+b)))`

Options

A. `tan^(-1)((a - b)/(2 sqrt(ab)))`
B. `tan^(-1)((a + b)/(2 sqrt(ab)))`
C. `tan^(-1)((2ab)/(sqrt(a-b)))`
D. `tan^(-1)((2ab)/(sqrt(a+b)))`

Answer

A

Explanation

Let `P (a cos theta, a sin theta)` be a point on the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` and `Q (a cos theta, a sin theta)` be the corresponding point on the auxiliary circle `x^(2) + y^(2) = a^(2)`. The equations of tangents at P and Q to the respective curves are
`x/a cos theta + y/b sin theta = 1`
and , `x cos theta + y sin theta = a` respectively
Let `alpha` be the acute angle between these tangents. Then,
`tan alpha = |(-cot theta + b/a cot theta)/(1 + b/a cot^(2)theta)|`
`tan alpha = |(a - b)/(a tan theta + b cot theta)|`
Now, `AM ge GM`
`rArr (a tan theta + b cot theta)/(2) ge sqrt(a tan theta xx b cot theta)`
`rArr (a tan theta + b cot theta)/(2) ge sqrt(ab) rArr a tan theta + b cot theta ge 2 sqrt(ab)`
`therefore tan alpha le |(a - b)/(2 sqrt(ab))|`
Hence, the greatest value of alpha is `tan^(-1)((a - b)/(2sqrt(ab)))`