Find a point on the curve `y=3x^(2)-2x - 4 ` at which the tangent is parallel to the line `10x-y+7=0`.

Options

A. `(2, 4)`
B. `(1, 4)`
C. `(-2, 4)`
D. `(2, -4)`

Answer

A

Explanation

Given line is ` 10x-y+7=0`
Slope of this line`m_(1)=-10/(-1)=10`
Equation of curve ` y=3x^(2)- 2x-4 …(1)`
` rArr (dy)/(dx) = 6x-2`
Let the point be `(x_(1),y_(1))`.
Slope of tangent at point `(x_(1), y_(1))`
`m= 6x_(1) - 2`
For parallel line
`m=m_(1)`
`rArr 6x_(1) - 2 =10 rArr x_(1) = 2`
The point `(x_(1), y_(1))` lies on curve (1).
`:. y_(1) = 3x_(1)^(2) - 2x_(1) - 4 = 3(2)^(2) - 2 (2) - 4`
` = 12- 4 - 4 = 4`.
`rArr ` The required point is (2, 4).