Question

For what values of
a
and
b
is (2, 2.5) is an inflection point of the curve
x^2 y + ax + by = 0
? What additional inflection points does the curve have?

Answer 1

first off, if (2,2.5) is an inflection point, then it is on the graph. That means

x^2 y + ax + by = 0
4*2.5 + 2a + 2.5b = 0
10 + 2a + 2.5b = 0
5/2 b = -(10+2a)
b = -2/5 (10+2a)

That gives us

x^2 y + ax - 2/5 (10+2a) y = 0
y = -ax/(x^2 - 2/5 (10+2a))
= 5ax/(4(a+5)-5x^2)
y' = 5a(4a+5(x^2+4))/(4(a+5)-5x^2)^2
y" = 50ax(12(a+5)+5x^2)/(4(a+5)-5x^2)^3

y"=0 when
a=0
b = -2/5 (10+2a) = -4

But with a=0, that gives us

x^2y - 4y = 0
y(x^2-4) = 0
which is just straight lines.

Otherwise, we can have

12(a+5)+5x^2=0
12(a+5)+20=0
a+5 = -5/3
a = -20/3
b = -2/5 (10+2(-20/3)) = 4/3

so, that gives us

x^2 y - 20/3 x + 4/3 y = 0
y = (20/3 x)/(x^2+4/3) = 20x/(3x^2+4)

The graph at

http://www.wolframalpha.com/input/?i=%2820%2F3+x%29%2F%28x^2%2B4%2F3%29

confirms this. You can also check the other points of inflection.

Answer 2

thank you for your help, i am confused as to what happened to the ys in this

x^2 y + ax - 2/5 (10+2a) y = 0
y = -ax/(x^2 - 2/5 (10+2a))

how did you get y by itself

Answer 3

never mind, did not see the other y

Answer 4

x^2 y + ax - 2/5 (10+2a) y = 0
2xy + x^2y' + a - 2/5 (10+2a)y' = 0
y'(x^2-2/5 (10+2a)) = -(2xy+a)
y' = 5(a+2xy)/(4(a+5)-5x^2)

y" =

5(2y + 2xy')(4(a+5)-5x^2) - 5(a+2xy)(-10x)
-----------------------------
(4(a+5)-5x^2)^2

5(2y + 2x(5(a+2xy)/(4(a+5)-5x^2)))(4(a+5)-5x^2) - 5(a+2xy)(-10x)
-----------------------------
(4(a+5)-5x^2)^2

20a(5x+2y)+50y(3x^2+4)
----------------------------
(4(a+5)-5x^2)^2

That is zero when the numerator is zero:

20a(5x+2y)+50y(3x^2+4) = 0
20a(5*2+2*2.5)+50*2.2(3*4+4) = 0
1760+300a = 0
a = -88/15

Hmmm. Should have gotten -100/15. Better check my math.