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. -6
D. none of these

Answer

B

Explanation

For the point (2, -1) on the curve `x=t^(2)+3t-8,y=2t^(2)- 2t -5` we have
`t^(2)+3t-8=2 and 2t^(2)-2t-5=-1`
`rArr t^(2)+3t-10=0 and 2t^(2)-2t-4=0`
`rArr (t+5)(t-2)=0 and (t-2)(t+1)=0`
`rArr t=2`
Now, `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(4t-2)/(2t+3) rArr ((dy)/dx)_(t=2)=(4xx2-2)/(2xx2+3)=(6)/(7)`