4.63 Find all values of \( x \) that satisfy the equation
$$\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$$

where \( p \) is a real parameter.
[Solve] $\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$
Rearrange and square both sides to get
$$x^{2}-p=x^{2}+4\left(x^{2}-1\right)-4 x \sqrt{x^{2}-1}$$

which simplifies to $4\left(x^{2}-1\right)+p=4 x \sqrt{x^{2}-1}$.
Square both sides again to get
$$16\left(x^{2}-1\right)^{2}+8 p\left(x^{2}-1\right)+p^{2}=16 x^{2}\left(x^{2}-1\right)$$

Rearranging gives $\quad 8(2-p) x^{2}=(p-4)^{2}$.
Now, we discuss the parameter $p$.
(1) When $p=2$, the equation (2) becomes
$$0 \cdot x^{2}=4$$

Clearly, there is no solution. Thus, the original equation has no solution.
(2) When $p>2$, for any real number $x$, we have
$$8(2-p) x^{2} \leqslant 0$$

while the right side of (2) is $\quad(p-4)^{2} \geqslant 0$.
Thus, only when $p=4$, equation (2) has a solution: $x=0$.
However, substituting $x=0$ into the original equation (1), the left side is undefined, so $x=0$ is not a solution to the original equation. Therefore, when $p>2$, the original equation has no solution.
(3) When $p<2$, we have $8(2-p) x^{2} > 0$, so
$$x=\frac{4-p}{2 \sqrt{2(2-p)}}$$

Substituting (3) into the original equation (1) for verification, we get
$$\frac{|3 p-4|}{2 \sqrt{2(2-p)}}+\frac{2|p|}{2 \sqrt{2(2-p)}}=\frac{4-p}{2 \sqrt{2(2-p)}}$$

which simplifies to
$$|3 p-4|+2|p|=4-p$$

We discuss $p$ in different cases.
When $p>\frac{4}{3}$, the original equation has no solution.
When $0 \leqslant p \leqslant \frac{4}{3}$, the original equation has a unique solution:
$$x=\frac{4-p}{2 \sqrt{2(2-p)}}.$$