A tangent to the ellipse `4x^2 +9y^2 =36` is cut by the tangent at the extremities of the major axis at T and `T^1`, the circle on `T T^1` as diameter passes through the point

Options

A. `(-sqrt5, 0)`
B. `(sqrt5, 1)`
C. `(0, 0)`
D. `(3, 2)`

Answer

C

Explanation

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).