Question: My method is to find the equation of the chord wrt to the point $P$ and the midpoint, say $O(\alpha,\beta)$ and then compare them.

Equation of the the chord wrt to $P$ is$^1$:
$$T=0 \text{ where } T=12x+20y-1.$$

So the equation of the chord is: $$12x+20y-1=0\tag1$$

The equation of the chord wrt to $O$ is$^2$

$$T^{'}=S\_1 \text{ where } T^{'}=3\alpha x+4\beta y-1 \text{ and } S\_1=3\alpha^2+4\beta ^2-1. $$

So the equation of the chord is: $$3\alpha x+4\beta y-1-3\alpha^2-4\beta ^2+1=0\tag2$$

Now since $(1) \text{ and } (2) $ are equations of same line then they are equivalent then we have on comparing $$ O(\alpha,\beta)=O(4,5). $$

Which is wrong and absurd, but why? There seems to be no mistake in this method. The answer is $O(\frac{1}{37},\frac{5}{148}).$

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The diagram of the question is [here](https://i.stack.imgur.com/3eDzh.png).

1: $T=0$ is true for all conics, and hence for [circles](https://www.askiitians.com/iit-jee-coordinate-geometry/chord-of-contact.aspx) as well.

2: $T=S\_1$ is true for all conic curves in general, but I found this shown [online](https://math.stackexchange.com/questions/3249671/equation-of-chord-of-a-parabola-whose-midpoint-is-given) for parabolas only.

Answer: 
**Hint:** $\dfrac{(x+h)\sqrt[3]{x+h} - x\sqrt[3]{x}}{h}= x\cdot\dfrac{\sqrt[3]{x+h} - \sqrt[3]{x}}{h}+\sqrt[3]{x+h}$. From this you can simplify the factor next to $x$ of the first term using the identity: $a - b = \dfrac{a^3 - b^3}{a^2+ab+b^2}$
You seemed to have trapped yourself by writing $f(x)=x^{4/3}$ in the form of a product that makes the task less transparent. Doing so leads more naturally into the need for the basic technique of the product rule of "plus then minus the cross term".

Namely, take $F(x) = x$ and $G(x) = \sqrt[3]{x}$ such that $f(x)=F(x)G(x)$.

$$\begin{align}
&F(x+h)G(x+h)-F(x)G(x) \\
&= F(x+h)G(x+h) ~\mathbf{{}- F(x+h)G(x) + F(x+h)G(x) }~- F(x)G(x) \\
&= F(x+h)\bigl( G(x+h)-G(x) \bigr) - \bigl(F(x+h) - F(x) \bigr)G(x)
\end{align}$$

Then you have two separate simpler derivatives that becomes the product rule $(FG)'=FG'+F'G$.