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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 throughA. `(3,0)`B. `(0,0)`C. `(0,3)`D. `(4,0)` |
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Answer» Correct Answer - 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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