4.20. Given the equation of a line lying in the XY plane, $4 x^{2}+p y^{2}=1$ and the coordinates of point $A(3 / 2,1,1)$. It is required: 1) to derive the equation of the surface formed by rotating this line around the $O X$ axis; 2) to select the value of the parameter $p$ so that point $A$ lies on the surface of rotation.
Solution.

1) Since the line rotates around the axis $O X$, to obtain the equation of the surface of revolution, it is necessary to replace $y$ with $\pm \sqrt{y^{2}+z^{2}}$ in the relation $4 x^{2}+p y^{2}=1$. We have

$$
4 x^{2}+p\left( \pm \sqrt{y^{2}+z^{2}}\right)^{2}=1, \quad 4 x^{2}+p\left(y^{2}+z^{2}\right)=1
$$

2) Substitute the coordinates of point $A$ for $x, y, z$ and find the value of the parameter $p$:

$$
\mathbf{4} \cdot(3 / 2)^{2}+p\left(1^{2}+1^{2}\right)=1, p=-4
$$

The equation of the surface of revolution, after replacing $p$ with $-4$, takes the form

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

The last equation can be rewritten as:

$$
y^{2} / 0.25+z^{2} / 0.25-x^{2} / 0.25=-1
$$

We have obtained the equation of a hyperboloid of two sheets of revolution.