241. At which points of the curve $x=t-1, y=t^{3}-12 t+1$ is the tangent parallel to: 1) the $Ox$ axis; 2) the line $9 x+y+3=0$?
Solution. Here we use the condition of parallelism of lines, which consists in the equality of their angular coefficients.

Let's find the derivative of $y$ with respect to $x$ from the equations of the curve:

$$
y^{\prime}=\frac{d y}{d t}: \frac{d x}{d t}=\frac{3 t^{2}-12}{1}=3 t^{2}-12
$$

This derivative represents the angular coefficient of the tangent to the given curve at any of its points.

1) By equating $y^{\prime}$ to the angular coefficient of the $O x$ axis, which is zero, we get $3 t^{2}-12=0 ; t^{2}=4 ; t= \pm 2$.

Substituting these values of the parameter $t$ into the given equations of the curve, we find the coordinates of those points where the tangent is parallel to the $O x$ axis: $(1 ;-15) ;(-3 ; 17)$.

2) By equating $y^{\prime}$ to the angular coefficient of the given line, which is -9, we get $3 t^{2}-12=-9 ; t^{2}=1 ; t= \pm 1$.

Using the found values of the parameter $t$ from the equations of the curve, we determine the coordinates of the desired points where the tangent to the curve is parallel to the given line: $(0 ;-10),(-2 ; 12)$.

242*. Form the equations of the tangents to the parabola $y=x^{2} -4 x+1$, passing through a point not lying on it: 1) $O(0 ; 0)$; 2) $A(1 ; 1)$.

Solution. The equation of the tangent to the given parabola has the general form

$$
y-y_{0}=\left(x^{2}-4 x+1\right)_{0}^{\prime}\left(x-x_{0}\right)
$$

or

$$
y-\left(x_{0}^{2}-4 x_{0}+1\right)=\left(2 x_{0}-4\right)\left(x-x_{0}\right)
$$

where $(x, y)$--is the current point on the tangent;

$\left(x_{0}, y_{0}\right)$ - the unknown point of tangency.

1) Since the tangent passes through point $O$, then

$$
0-\left(x_{0}^{2}-4 x_{0}+1\right)=\left(2 x_{0}-4\right)\left(0-x_{0}\right)
$$

Solving this quadratic equation, we find two values for the abscissa of the point of tangency $x_{0}$: $x_{0}= \pm 1$, and from this the equations of the two tangents: $2 x+y=0$ and $6 x+y=0$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-084.jpg?height=521&width=246&top_left_y=723&top_left_x=102)

Fig. 41

2) For point $A$, the same reasoning leads to the quadratic equation $x_{0}^{2}-2 x_{0}+4=0$, the roots of which are complex. Therefore, no tangent can be drawn from point $A$ to the given parabola.

The obtained results have a simple geometric meaning: from each point belonging to the external region of the parabola, two tangents can be drawn to it, and from a point belonging to its internal region, none (Fig. 41).

In the general case, the problem of drawing tangents to a curve $y=f(x)$ through a point $(a, b)$ not lying on this curve is solved in the same way, based on the general equation of the tangent

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

This problem has as many solutions as the number of real roots of the equation

$$
b-f\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)\left(a-x_{0}\right)
$$

In problems 243-248, find the equations of the tangents and normals to the given curves at the specified points and construct the curves, tangents, and normals.