Determine a and b so that the point (2, dfrac52) is

Determine $a$ and $b$ so that the point $\left(2,\,\dfrac52\right)$ is an inflection point of the curve $x^{2}y+ax+by = 0$. Has the curve any further inflection points?

Step 1: Solve for $y$

\[ y\bigl(x^{2}+b\bigr) = -ax \;\Longrightarrow\; y = \frac{-ax}{x^{2}+b}. \tag{1} \]

Step 2: The point condition

\[ \frac52 = \frac{-2a}{4+b}. \tag{2} \]

Step 3: The derivatives

\[ y' = \frac{-a\bigl(x^{2}+b\bigr)+2ax^{2}}{\bigl(x^{2}+b\bigr)^{2}} = \frac{a\bigl(x^{2}-b\bigr)}{\bigl(x^{2}+b\bigr)^{2}}, \] and differentiating again, with the bracket $\bigl(x^{2}+b\bigr)-2\bigl(x^{2}-b\bigr) = 3b-x^{2}$, \[y'' = \frac{2ax\bigl(3b-x^{2}\bigr)}{\bigl(x^{2}+b\bigr)^{3}}. \tag{3}\]

Step 4: The inflection condition

Setting $x = 2$ in (3), with $a\neq0$ (otherwise $y\equiv0$) and $4+b\neq0$: \[\frac{4a\bigl(3b-4\bigr)}{\bigl(4+b\bigr)^{3}} = 0 \;\Longrightarrow\; 3b = 4 \;\Longrightarrow\; b = \frac43. \tag{4}\]

Step 5: And then $a$

Substituting $b = \tfrac43$ into (2): \[\frac52 = \frac{-2a}{\tfrac{16}{3}} = -\frac{3a}{8} \;\Longrightarrow\; a = -\frac{20}{3}.\]

Step 6: The curve, and all its inflections

With these values $3b = 4$, so (3) becomes

\[ y'' = \frac{2a\,x\bigl(4-x^{2}\bigr)}{\left(x^{2}+\tfrac43\right)^{3}} = \frac{\tfrac{40}{3}\,x\bigl(x^{2}-4\bigr)}{\left(x^{2}+\tfrac43\right)^{3}}, \qquad y = \frac{\tfrac{20}{3}x}{x^{2}+\tfrac43}. \tag{5} \]

The denominator is positive, so the sign of $y''$ is that of $x\bigl(x-2\bigr)\bigl(x+2\bigr)$: $x\lt -2:\ -; \qquad -2\lt x\lt 0:\ +; \qquad 0\lt x\lt 2:\ -; \qquad x\gt 2:\ +$.

All three roots are simple, so all three are inflection points: \[y(2) = \frac{\tfrac{40}{3}}{\tfrac{16}{3}} = \frac52, \qquad y(0) = 0, \qquad y(-2) = -\frac52.\]

Step 7: Verification

\[y = \frac{20x}{3x^{2}+4}:\quad y(2) = \frac{40}{16} = 2.5 \;\;\checkmark; \qquad y(1) = \frac{20}{7} = 2.857\]

\[y''(1) = \frac{\tfrac{40}{3}\cdot(-3)}{\left(\tfrac73\right)^{3}} = -3.149\lt 0; \qquad y''(3) = \frac{\tfrac{40}{3}\cdot3\cdot5}{\left(\tfrac{31}{3}\right)^{3}} = 0.1813\gt 0 \;\;\checkmark\]

\[\text{The function is odd, so the inflections come in the symmetric pair } \pm2 \text{ together with the origin} \;\;\checkmark\]