240. Find the angles at which the following lines intersect:
1) the line $x+y-4=0$ and the parabola $2 y=8-x^{2}$;
2) the ellipse $x^{2}+4 y^{2}=4$ and the parabola $4 y=4-5 x^{2}$;
3) the sine curve $y=\sin x$ and the cosine curve $y=\cos x$.
Solution. 1) Solving the equations of the parabola and the line together, we find that they intersect at two points: $A(0 ; 4)$ and $B(2 ; 2)$, see Fig. 38.

Next, we find the derivative of $y$ with respect to $x$ from the equation of the parabola: $2 y^{\prime}=-2 x, y^{\prime}=-x$ and determine the slopes of the tangents to the parabola at points $A$ and $B$, as particular values of this derivative:

$$
\dot{y}_{A}^{\prime}=k_{A}=0 ; \quad y_{B}^{\prime}=k_{B}=-2
$$

The slope of the line is the same at all its points; for the given line, it is -1.

According to formula (2), we get

$$
\operatorname{tg} A=1, \quad A=45^{\circ} ; \quad \operatorname{tg} B=\frac{-1+2}{1+2}=\frac{1}{3}, \quad B \approx 18.5^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-082.jpg?height=286&width=320&top_left_y=628&top_left_x=204)

Fig. 37

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-082.jpg?height=261&width=305&top_left_y=654&top_left_x=763)

Fig. 38

2) Solving the equations of the curves together, we find their common points: $A(1.2 ; -0.8), B(0 ; 1)$, and $C(-1.2 ; -0.8)$, see Fig. 39. Then we determine the slopes $k_{1}$ and $k_{2}$ of the tangents at any point on the ellipse and the parabola as the derivatives of $y$ with respect to $x$ from their equations

$$
k_{1}=-\frac{x}{4 y} \quad \text { and } \quad k_{2}=-\frac{5}{2} x
$$

Substituting the coordinates of point $A$, we get $k_{1}=\frac{3}{8}$ and $k_{2}=-3$. Therefore, at point $A$:

$$
\operatorname{tg} \varphi=\frac{\frac{3}{8}+3}{1-\frac{9}{8}}=-27 ; \quad \varphi \approx 92^{\circ}
$$

The curves intersect at the same angle at point $C$ due to their symmetry with respect to the $O y$ axis.

At point $B$, we have: $k_{1}=k_{2}=0$, so at point $B$ the curves have a common tangent, i.e., they touch each other. At this point, the angle between the curves is zero.

3) The abscissas of the points of intersection of the curves (Fig. 40) are determined by the equation $\sin x=\cos x$, solving which, we get

$$
x=\frac{\pi}{4}+n \pi(n=0, \pm 1, \pm 2 \ldots)
$$

By differentiation, we find the slopes of the tangents to the sine and cosine curves: $k_{1}=\cos x ; k_{2}=-\sin x$.

The desired angle between the curves is determined by the general formula (2)

$$
\operatorname{tg} \varphi=\frac{\cos x+\sin x}{1-\cos x \sin x}= \pm \frac{\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}}{1-\frac{1}{2}}= \pm 2 \sqrt{2}
$$

The positive sign corresponds to the acute angle $\varphi \approx 70.5^{\circ}$. The negative sign corresponds to the obtuse, supplementary angle $\varphi_{1} \approx 109.5^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-083.jpg?height=288&width=353&top_left_y=674&top_left_x=162)

Fig. 39

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-083.jpg?height=194&width=489&top_left_y=749&top_left_x=626)

Fig. 40