Answer
$a=-\frac{20}{3}$
$b = \frac{4}{3}$
The other inflection points are $(0,0)$ and $(-2, -2.5)$
Work Step by Step
$x^2y+ax+by=0$
$y(x^2+b) = -ax$
$y = -\frac{ax}{x^2+b}$
We can find $y'$:
$y' = -\frac{a(x^2+b)-(ax)(2x)}{(x^2+b)^2}$
$y' = -\frac{ax^2+ab-2ax^2}{(x^2+b)^2}$
$y' = \frac{ax^2-ab}{(x^2+b)^2}$
We can find $y''$:
$y'' = \frac{(2ax)(x^2+b)^2-(ax^2-ab)(2)(x^2+b)(2x)}{(x^2+b)^4}$
$y'' = \frac{(2ax)(x^2+b)-(ax^2-ab)(2)(2x)}{(x^2+b)^3}$
$y'' = \frac{2ax^3+2abx-4ax^3+4abx}{(x^2+b)^3}$
$y'' = \frac{6abx-2ax^3}{(x^2+b)^3}$
To find the points of inflection, we can find values of $x$ such that $y'' = 0$:
$y'' = \frac{6abx-2ax^3}{(x^2+b)^3} = 0$
$6abx-2ax^3 = 0$
$-2ax(x^2-3b) = 0$
$x = 0$ or $x^2 = 3b$
$x = 0$ or $x = \pm \sqrt{3b}$
It is given that $(2,2.5)$ is an inflection point.
$x = \sqrt{3b} = 2$
$3b = 4$
$b = \frac{4}{3}$
We can find $a$:
$x^2y+ax+by=0$
$(2)^2(2.5)+2a+(\frac{4}{3})(2.5)=0$
$10+2a+\frac{10}{3}=0$
$2a=-10-\frac{10}{3}$
$2a=-\frac{40}{3}$
$a=-\frac{20}{3}$
We can find the other inflection points:
When $x = 0$:
$x^2y+ax+by=0$
$(0^2)y+a(0)+\frac{4}{3}y=0$
$0+0+\frac{4}{3}y=0$
$y = 0$
When $x = -2$:
$x^2y+ax+by=0$
$(-2)^2y+(-\frac{20}{3})(-2)+\frac{4}{3}y=0$
$4y+\frac{40}{3}+\frac{4}{3}y=0$
$12y+40+4y=0$
$16y = -40$
$y = -\frac{40}{16}$
$y = -2.5$
The other inflection points are $(0,0)$ and $(-2, -2.5)$
