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A tangent to the ellipse Ax^(2)+9y^(2)=3...

A tangent to the ellipse `Ax^(2)+9y^(2)=36` is cut by the tangent at the extremities of the major axis at T and T`. The circle TT' as diameter passes through the point

A

`(-sqrt5, 0)`

B

`(sqrt5, 1)`

C

`(0, 0)`

D

`(3, 2)`

Text Solution

Verified by Experts

The correct Answer is:
C

Let `P(3 cos theta, 2 sin theta)` be a point on `4x^(2) + 9y^(2) = 36`. The equation of the tangent at P is
`2x cos theta + 3y sin theta = 6`
This meets the coordinate axes at `T (3 sec theta, o) and T' (0, 2 " cosec " theta).`
The equation of the circle with TT' as diameter is `(x - 3 sec theta) (x - 0) + (y - 0) (y - 2 " cosec " theta) = 0`
or, `x^(2) + y^(2) - 3x sec theta - 2y "c cosec " theta = 0`
Clearly, it passes through (0, 0).
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