Write the coordinates of the point

Solution:

Let $(a, b)$ be the required coordinate.

Given that the tangent to the curve $y=2 x^{2}-x+1$ is parallel to the line $y=3 x+9$.

Slope of the line $=3$

$\because$ the point lies on the curve

$\Rightarrow \mathrm{b}=2 \mathrm{a}^{2}-\mathrm{a}+1 \ldots$  (1)

Now, $y=2 x^{2}-x+1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{x}-1$

Now value of slope at $(a, b)$

$\Rightarrow \frac{d y}{d x}=4 a-1$

Given that Slope of tangent $=$ Slope of line

$\Rightarrow 4 a-1=3$

$\Rightarrow 4 a=4$

$\Rightarrow a=1$

From (1),

$b=2-1+1$

$\Rightarrow b=2$