Find the equation of the line tangent to the curve x=2 t^{3}-15 t^{2}+24 t+7, \quad y=t^{2}+t+1 \quad \text { at } \quad t=2.
Chapter 2
Q. 2.2
Step-by-Step
Verified Solution
Here f_{1}^{\prime}(2)=\left.\left(6 t^{2}-30 t+24\right)\right|_{t=2}=-12 \text { and } f_{2}^{\prime}(2)=\left.(2 t+1)\right|_{t=2}=5.
When t=2, x=11 and y=7. Then using (4),
f_{2}^{\prime}\left(t_{0}\right)\left(x-x_{0}\right)-f_{1}^{\prime}\left(t_{0}\right)\left(y-y_{0}\right)=0we obtain
5(x-11)+12(y-7)=0, \quad \text { or } \quad 5 x+12 y=139.