Write the coordinates of the point at which the tangent to the curve y = 2x2 – x + 1 is parallel to the line y = 3x + 9.
Let (a, b) be the required coordinate.
Given that the tangent to the curve y = 2x2 – x + 1 is parallel to the line y = 3x + 9.
Slope of the line = 3
∵ the point lies on the curve
⇒ b = 2a2 – a + 1 … (1)
Now, y = 2x2 – x + 1
Now value of slope at (a, b)
Given that Slope of tangent = Slope of line
⇒ 4a – 1 = 3
⇒ 4a= 4
⇒ a = 1
From (1),
b = 2 – 1 +1
⇒ b = 2