For the family y = (x²+1)/(x²- alpha), alpha < 0, find

For the family $\displaystyle y = \frac{x^{2}+1}{x^{2}-\alpha}$, $\alpha\lt 0$, find the curve on which all the inflection points lie.

Step 1: A convenient form

Since $\alpha\lt 0$, the denominator $x^{2}-\alpha = x^{2}+|\alpha|\gt 0$ never vanishes and the function is defined on all of $\mathbb{R}$. Dividing out, \[y = \frac{\bigl(x^{2}-\alpha\bigr)+\bigl(1+\alpha\bigr)}{x^{2}-\alpha} = 1+\frac{1+\alpha}{x^{2}-\alpha}. \tag{1}\]

Step 2: Two derivatives

\[ y' = \bigl(1+\alpha\bigr)\cdot\frac{-2x}{\bigl(x^{2}-\alpha\bigr)^{2}}, \] and differentiating once more, the factor $\bigl(x^{2}-\alpha\bigr)$ cancelling once: \[y'' = \frac{2\bigl(1+\alpha\bigr)\left[-\bigl(x^{2}-\alpha\bigr)+4x^{2}\right]}{\bigl(x^{2}-\alpha\bigr)^{3}} = \frac{2\bigl(1+\alpha\bigr)\bigl(3x^{2}+\alpha\bigr)}{\bigl(x^{2}-\alpha\bigr)^{3}}. \tag{2}\]

Step 3: The inflections

For $\alpha\neq-1$ the constant $1+\alpha$ is non-zero and the denominator is positive, so

\[ y'' = 0 \;\Longleftrightarrow\; 3x^{2}+\alpha = 0 \;\Longleftrightarrow\; x^{2} = -\frac{\alpha}{3}, \qquad x = \pm\sqrt{-\frac{\alpha}{3}}, \tag{3} \] real because $\alpha\lt 0$; the roots are simple, so both are genuine inflections. (For $\alpha = -1$ the function is the constant $1$ and there is nothing to find.)

Step 4: The height

Substituting $x^{2} = -\dfrac{\alpha}{3}$ into (1): \[y = 1+\frac{1+\alpha}{-\dfrac{\alpha}{3}-\alpha} = 1-\frac{3\bigl(1+\alpha\bigr)}{4\alpha} = \frac{4\alpha-3-3\alpha}{4\alpha} = \frac{\alpha-3}{4\alpha}. \tag{4}\]

Step 5: Eliminate $\alpha$

From (3), $\alpha = -3x^{2}$; putting this into (4): \[y = \frac{-3x^{2}-3}{-12x^{2}} = \frac{x^{2}+1}{4x^{2}} \;\Longleftrightarrow\; 4x^{2}y = x^{2}+1 \;\Longleftrightarrow\; x^{2}\bigl(4y-1\bigr) = 1.\]

Step 6: Verification

\[\alpha = -3:\ x = \pm1,\ y = \frac{-3-3}{-12} = 0.5,\ \text{and } x^{2}\bigl(4y-1\bigr) = 1\cdot1 = 1 \;\;\checkmark\]

\[\alpha = -12:\ x = \pm2,\ y = 0.3125,\ \text{and } 4\bigl(1.25-1\bigr) = 1 \;\;\checkmark\]

\[y-\frac14 = \frac{\alpha-3}{4\alpha}-\frac14 = \frac{-3}{4\alpha}\gt 0 \text{ for } \alpha\lt 0,\ \text{consistent with } 4y-1 = \frac{1}{x^{2}}\gt 0 \;\;\checkmark\]