Saved Bookmarks
| 1. |
On the ellipse `4x^2+9y^2=1,`the points at which the tangents are parallel to the line `8x=9y`are`(2/5,1/5)`(b) `(-2/5,1/5)``(-2/5,-1/5)`(d) `(2/5,-1/5)`A. `((2)/(5), (1)/(5))`B. `(-(2)/(5), (1)/(5))`C. `(-(2)/(5), - (1)/(5))`D. `((2)/(5), - (1)/(5))` |
|
Answer» Correct Answer - B::D Given, ` 4x ^(2 ) + 9y^(2) = 1 " " `… (i) On differentiating w.r.t. x, we get `" " 8x + 18y (dy)/(dx) =0` `rArr " " (dy)/(dx) =- ( 8x)/( 18y ) =- ( 4x )/( 9y)` The tangent at point `(h, k)` will be parallel to ` 8x = 9y`, then `" " - ( 4h)/( 9k) = (8)/(9)` `rArr " " h = - 2k ` Point `(h, k)` also lies on the ellipse. `therefore " " 4h^(2) + 9k^(2) = 1" " `... (ii) On putting value of h in Eq. (ii), we get `" " 4 (-2 k)^(2) + 9k ^(2) = 1 ` `rArr " " 16 k ^(2) = 9 k^(2) = 1 ` `rArr " " 25 k^(2) = 1 ` `rArr " " k ^(2) = (1)/( 25)` `rArr " " k = pm (1)/(5)` Thus, the point, where the tangents are parallel to `8x = 9y ` are `(-(2)/(5), (1)/(5)) and ((2)/(5), - (1)/(5))`. Therefore, options (b) and (d) are the answers. |
|