Given equation of curve is x = t2 + 3t – 8, y = 2t2 – 2t – 5
x = t2 + 3t – 8
Differentiating on both sides with respect to t, we get
y = 2t2 – 2t – 5
Differentiating on both sides with respect to t, we get
We know derivative of a constant is 0, so above equation becomes
As given the curve passes through the point (2,-1), substituting these values in given curve equation we get
x = t2 + 3t – 8
⇒ 2= t2 + 3t – 8
⇒ t2 + 3t – 8-2=0
⇒ t2 + 3t – 10=0
Splitting the middle term we get
⇒ t2 + 5t-2t – 10=0
⇒ t(t+ 5) -2(t+5)=0
⇒ (t+ 5) (t-2)=0
⇒ t+5=0 or t-2=0
⇒ t=-5or t=2……….(iii)
y = 2t2 – 2t – 5
⇒ -1=2t2 – 2t – 5
⇒ 2t2 – 2t – 5+1=0
⇒ 2t2 – 2t – 4=0
Taking 2 common we get
⇒ t2 – t – 2=0
Splitting the middle term we get
⇒ t2 – 2t +t– 2=0
⇒ t(t– 2)+1(t– 2)=0
⇒ (t– 2)(t+1)=0
⇒ (t– 2)=0 or (t+1)=0
⇒ t=2 or t=-1……..(iv)
So from equation (iii) and (iv), we can see that 2 is common
So t=2
So the slope of the tangent at t=2 is given by
Therefore, the slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is 6/7.
So the correct option is option B.