How to show that the curve $ay^2=x(x-a)(x-b)$ has two and only two points of inflexion?

Question

Show that the curve $$ ay^{2}=x(x-a)(x-b) $$ has two and only two points of inflexion
This is a question that I have stumbled upon while going through the book
Differential Calculcus by Shanti Narayan and Dr P.K. Mittal.
How I approached the problem was to change the equation to
$$ y= pm sqrt{frac{1}{a}x(x-a)(x-b)} $$
Then I went on to find the 2nd derivative of y and equated it to 0 which gave me the biquadratic equation
$$ 3x^{4} - 4left(a+bright)x^{3} + 6abx^{2} - a^{2}b^{2} = 0 $$
After this I got stuck. I tried out many ways like the Descarte's rule of signs, but none of them gave me the actual solution.
It would be very helpful if you could help me out with this problem.

Maths Wizzard · Answer

The way I will present here I think is the easiest to grasp. As you have correctly calculated we have
$$g(x)=f''(x)= 3x^4 − 4(a+b)x^3 + 6abx^2 − a^2b^2  .$$
We aim to show there are two unique solutions. We will try to get a sense for the graph by trying to figure out where the turning points of $g$ are at. We have
$$g'(x)=12x^3-12x^2(a+b)+12abx=12x(x^2-x(a+b)+ab)$$ the solutions of which are $x=0$ and $x=a$ and $x=b$.  We have $g(0)=-a^2b^2$ and  $g(a)=-a^2(a-b)^2$ and $g(b)=-b^2(a-b)^2$. Thus all turnining points have negative coordinates. Taking into account that $lim_{xrightarrow ± infty} g(x) $ tends to positive infinity in both cases, we have the following graph
Graph.
From it, we clearly see that there are two solutions

How to show that the curve $ay^2=x(x-a)(x-b)$ has two and only two points of inflexion?

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