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: $a - b = \dfrac{a^3 - b^3}{a^2+ab+b^2}$
If you multiply both the numerator and the denominator of that fraction by$$(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2,$$then the limit becomes\begin{multline}\lim\_{h\to0}\frac{(x+h)^3(x+h)-x^4}{h\bigl((x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2\bigr)}=\\=\lim\_{h\to0}\frac{6hx^2+4h^2x+h^3+4x^3}{(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2}.\end{multline}Can you take it from here?