KCET 2016 Math Solved Paper

Given that,x=t2+3t−8→ (1)
y=2t2−2t−5 → (2)
At point (2, −1), Eq. (2) becomes
−1=2t2−2t−5
⇒2t2−2t−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+3t−8
⇒t2+3t−10=0
⇒(t+5)(t−2)=0
⇒t=2,−5 → (4)
From Eqs. (3) and (4), we have common value of t = 2.
Now, ‌

dy

dt

=4t−2 and ‌

dx

dt

=2t+3
So, slope of the tangent to the curve is
‌

dy

dx

=‌

4t−2

2t+3

⇒(‌

dy

dx

)t=2=‌

4(2)−2

(2)+3

=‌

6

7