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The tangent at any point on the ellipse ...

The tangent at any point on the ellipse `16x^(2)+25y^(2) = 400` meets the tangents at the ends of the major axis at `T_(1)` and `T_(2)`. The circle on `T_(1)T_(2)` as diameter passes through

A

`(3,0)`

B

`(0,0)`

C

`(0,3)`

D

`(4,0)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(x^(2))/(5^(2)) +(y^(2))/(4^(2)) =1`
Any tangent to the ellipse is `(x cos theta)/(5) + (y sin theta)/(4) =1`
This meets `x = a =5` at `T_(1) {5,(4)/(sin theta) (1-cos theta)}`
`= {5,4 tan.(theta)/(2)}` and meets `x =- a =- 5` at
`T_(2) {-5,(4)/(sin theta) (1+cos theta)} = {-5,4 cot.(theta)/(2)}`
The circle on `T_(1),T_(2)` as diameter is
`(x-5) (x+5) + (y-4tan.(theta)/(2)) = 0`
`x^(2) + y^(2) - 4y (tan.(theta)/(2)+cot.(theta)/(2)) -25 +16 =0`
This is obviously satisfied by (3,0).
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