Conic Section Equation from Michael Spivak's Book So i've been reading Michael Spivak's Calculus lately and now i feel im stuck in his conic section equation, page 81.

What i dont understand is, how can the first equation becomes the second? After squaring it, i got different coefficient especially in the x² part.
 A: There is indeed a typo. From:
$$M (\alpha x + \beta) + B = \pm C \sqrt{(\alpha x + \beta)^2 + y^2}$$
square both sides:
$$\left [M (\alpha x + \beta) + B \right ]^2 = \left [\pm C \sqrt{(\alpha x + \beta)^2 + y^2} \right]^2$$
Now on the left use $(a + b)^2 = a^2 + 2 a b + b^2$:
$$M^2(\alpha x + \beta)^2 + 2 B M (\alpha x + \beta) + B^2 = C^2 \left [(\alpha x + \beta)^2 + y^2 \right ]$$
Then subtract the right-hand side:
$$M^2(\alpha x + \beta)^2 + 2 B M (\alpha x + \beta) + B^2 - C^2 (\alpha x + \beta)^2 - C^2 y^2 = 0$$
Regroup the terms according to the powers of $(\alpha x + \beta)$:
$$(M^2 - C^2) (\alpha x + \beta)^2 + 2BM(\alpha x + \beta) + B^2 - C^2 y^2 = 0$$
Now we look at the coefficients of $x^2$ and $y^2$.
The coefficient of $x^2$ must come from the first term, because it's the only one in which a term containing $x$ is squared. Therefore it's $\alpha^2 (M^2 - C^2)$. Clearly, the coefficient of $y^2$ is $-C^2$. Therefore we can write
$$-C^2 y^2 + \alpha^2(M^2 - C^2) x^2 + Ex + F = 0$$
for some $E, F$ (notice that there are no terms with $y$). Changing signs, we get:
$$C^2 y^2 - \alpha^2 (M^2 - C^2) x^2 + E x + F = 0$$
or equivalently,
$$C^2 y^2 + \alpha^2 (C^2 - M^2) x^2 + E x + F = 0.$$
For the last equation, instead of dividing by $\alpha^2$, we can write:
$$C^2 y^2 + (C^2 - M^2) (\alpha x)^2 + \frac E \alpha (\alpha x) + F = 0$$
and so if we substitute $\alpha x$ with $x$ and let $G = \frac E \alpha$ and $H = F$, we get
$$C^2 y^2 + (C^2 - M^2) x^2 + G x + H = 0.$$
