If T is the length of the subtangent drawn at | vedclass

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(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$ Let $P(h, k)$ be a point on the curve. Differentiating equation $(i)$ with respect to

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$2 N$

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(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$ Let $P(h, k)$ be a point on the curve. Differentiating equation $(i)$ with respect to $x$,we get $6 y \frac{dy}{dx} = 12 x^2$,which implies $\frac{dy}{dx} = \frac{2 x^2}{y}$. At point $(h, k)$,the slope $m = \frac{2 h^2}{k}$. The length of the subtangent $T = \left| \frac{k}{m} \right| = \left| \frac{k}{2 h^2 / k} \right| = \frac{k^2}{2 h^2}$. Since $3 k^2 = 4 h^3$,we have $k^2 = \frac{4}{3} h^3$. Substituting this into $T$,we get $T = \frac{4 h^3}{3(2 h^2)} = \frac{2}{3} h$. The length of the subnormal $N = |k m| = \left| k \cdot \frac{2 h^2}{k} \right| = 2 h^2$. We need to find the relation between $T$ and $N$. From $N = 2 h^2$,we have $h^2 = \frac{N}{2}$,so $h = \sqrt{\frac{N}{2}}$. Then $T = \frac{2}{3} \sqrt{\frac{N}{2}}$. Squaring both sides,$T^2 = \frac{4}{9} \cdot \frac{N}{2} = \frac{2 N}{9}$. Thus,$9 T^2 = 2 N$,which can be written as $(3 T)^2 = 2 N$. Comparing this with $(\beta T)^2 = 2 N$,we find $\beta = 3$. Therefore,the correct option is $C$.

If $T$ is the length of the subtangent drawn at any point on the curve $3 y^2 = 4 x^3$ and $N$ is the length of the subnormal at the same point,then $(\beta T)^2 =$

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