376. On the curve $y=4 x^{2}-6 x+3$, find the point where the tangent line is: a) parallel to the line $y=2 x$; b) perpendicular to the line $y=x / 4$.
Solution. Let the desired point of tangency be ( $x_{0} ; y_{0}$ ). Then, as is known, the slope $k$ of the tangent line is equal to the value of the derivative at the point of tangency, i.e.,

$$
k=y_{x=x_{0}}^{\prime}=8 x-\left.6\right|_{x=x_{0}}=8 x_{0}-6=2\left(4 x_{0}-3\right)
$$

Considering this, let's examine each condition of the problem.

a) For the tangent line to be parallel to the line $y=2 x$, their slopes must be equal, i.e., $k=2$ or $2\left(4 x_{0}-\right.$ $-3)=2$. Solving the last equation for $x_{0}$, we get: $4 x_{0}-$ $-3=1 ; 4 x_{0}=4 ; x_{0}=1$. Substituting the found value of the abscissa of the desired point into the equation of the curve, we find the value of its ordinate: $y_{0}=4 \cdot 1^{2}-6 \cdot 1+3=1$. Thus, $(1 ; 1)$ is the desired point.

b) From analytic geometry, it is known that the slopes of mutually perpendicular lines are inverses in absolute value and opposite in sign. Since the slope of the line $y=x / 4$ is $1 / 4$, the slope $k$ of the desired tangent line is -4, and we have the equation $2\left(4 x_{0}-3\right)=-4$, from which $8 x_{0}=-4+6$, i.e., $x_{0}=1 / 4$. Accordingly, we find $y_{0}=4(1 / 4)^{2}-6(1 / 4)+3=7 / 4$. Therefore, the point ( $1 / 4 ; 7 / 4$ ) is the desired one.

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-223.jpg?height=202&width=402&top_left_y=322&top_left_x=197)

Fig. 105

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-223.jpg?height=340&width=402&top_left_y=184&top_left_x=645)

Fig. 106