Saved Bookmarks
| 1. |
The slope of the tangent to the curve `x=t^2+3t-8,``y=2t^2-2t-5`at the point `(2, 1)`is(A) `(22)/7` (B) `6/7` (C) `7/6` (D) `(-6)/7`A. `22/7`B. ` 6/7`C. ` 7/6`D. ` (-6)/7` |
|
Answer» Correct Answer - B Given, ` x= t^(2) + 3t - 8 and y = 2t^(2) - 2t - 5` …(1) ` rArr (dx)/(dt) = 2t+3 and (dy)/(dt) = 4t - 2` ` :. (dy)/(dx) = (dy)/(dt) xx (dt)/(dx) = (4t - 2)/(2t + 3)` Given point is (2, -1) at x = 2, from equation (1), ` 2 = t^(2) + 3t-8 rArr t^(2) + 3t - 10 = 0` ` rArr (t+5)(t-2)=0 rArr t=2 or t =- 5` at y =- 1 from equation (1), `-1=2t^(2)-2t-5 rArr 2t^(2)-2t - 4 = 0` ` rArr 2(t-2)(t+1)=0 rArr t = 2 or t =- 1` common value of t is 2. `:. ` slope of tangent of curve at point (2, -1) is `((dy)/(dx))_(t=2) = (4 xx 2-2)/(2 xx 2+ 3) = (8-2)/(4+3) = 6/7` . |
|