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Tangents are drawn to the ellipse (x^2)/...

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)))`

A

`tan^(-1)((a-b)/(2sqrt(ab)))`

B

`tan^(-1)((a+b)/(2sqrt(ab)))`

C

`tan^(-1)((2ab)/(2sqrt(a-b)))`

D

`tan^(-1)((2ab)/(2sqrt(a+b)))`

Text Solution

Verified by Experts


Tanent to the ellipse at `P( a cos alpha, b sin alpha)` is `(x)/(a) cos alpha+(y)/(b) isn alpha=1" "(1)`
Tangent to the circle at `Q(a cos, alpha, a sin alpha)` is ` cos ax+sin alphay=a " "(2)`
Now, the angle between the tangents is `theta`. Then, `tan theta=|(-(b)/(a)cotalpha-(-cot alpha))/(1+(-(b)/(a)cot alpha)(-cot alpha))|`
`=|(cos alpha(1-(b)/(a)))/(1+(b)/(a)cot^(2)alpha)|=|(a-b)/(a tan alpha+b cot alpha)|`
`=|(a-b)/((sqrt(atanalpha)-sqrt(bcotalpha))^(2)+2sqrt(ab))|`
Now, the greatest value of the above expression is `|(a-b)/(2sqrt(ab))|` when `sqrt(tan alpha)= sqrt (b tan alpha)`. Therefore,
`theta_("maximum")=tan^(-1)((a-b)/(2sqrt(ab)))`
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