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## Q. 2.2

Find the equation of the line tangent to the curve $x=2 t^{3}-15 t^{2}+24 t+7, \quad y=t^{2}+t+1 \quad \text { at } \quad t=2$.

## 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),

$f_{2}^{\prime}\left(t_{0}\right)\left(x-x_{0}\right)-f_{1}^{\prime}\left(t_{0}\right)\left(y-y_{0}\right)=0$

we obtain

$5(x-11)+12(y-7)=0, \quad \text { or } \quad 5 x+12 y=139$.