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. `-6/7`
D. `-6`

Answer

B

Explanation

Equation of curve is given by
`x=t^(2)+3t-8` and `y=2t^(2)-2t-5`.
`therefore (dx)/(dt)=2t+3` and `(dy)/(dx)=4t-2`
`rArr (dy)/(dx) = ((dy)/(dt))/(dx/dt)=(4t-2)/(2t+3)`..............(i)
Since, the curve passes throgh the point `(2,-1)`.
`therefore 2=t^(2)+3t-8`
and `-1=2t^(2)-2t-5`
`rArr t^(2)+3t-10=0`
and `t^(2)+5t-2t-10=0`
`rArr t(t+5)-2(t+5)=0`
`rArr (t-2)(t+5)=0`
and `(2t-4)(t+1)=0`
`rArr t=2,-5` and `t=-1,2`
`rArr t=2`
`therefore` Slope of tangent,
`(dy)/(dx)_("at" t=2)=(4 xx 2-2)/(2 xx 2+3)=6/7` [using Eq. (i)]