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`

Options

A. `22/7`
B. ` 6/7`
C. ` 7/6`
D. ` (-6)/7`

Answer

B

Explanation

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` .