The slope of the tangent to the curve ( x=t²+3 t-8, y=2 t²-2 t-5 ) at the point ( (2,-1) ) is

Step-by-step solution

Given that, x=t²+3 t-8 (1)y=2 t²-2 t-5 (2) At point (2,-1) , Eq. (2) becomes -1=2 t²-2 t-5 2 t²-2 t-4=0 t²-t-2=0 (t-2)(t+1)=0 t=2,-1 (3) Similarly, at point (2,-1) Eq. (1) becomes 2=t²+3 t-8 t²+3 t-10=0 (t+5)(t-2)=0 t=2,-5 (4) From Eqs. (3) and (4), we have common value of t=2 Now, d y d t =4 t-2 and d x d t =2 t+3 So, slope of the tangent to the curve is d y d x = 4 t-2 2 t+3 ( d y d x )_ t=2 = 4(2)-2 (2)+3 = 6 7