If the normal drawn at a point P on the curve | vedclass

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(A) Given the curve $3y = 6x - 5x^3$. Differentiating with respect to $x$,we get $3 \frac{dy}{dx} = 6 - 15x^2$,which simplifies to $\frac{dy}{dx} = 2

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$1$

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(A) Given the curve $3y = 6x - 5x^3$. Differentiating with respect to $x$,we get $3 \frac{dy}{dx} = 6 - 15x^2$,which simplifies to $\frac{dy}{dx} = 2 - 5x^2$. The slope of the normal at point $P(h, k)$ is $m_n = -\frac{1}{2 - 5h^2}$. The equation of the normal passing through $(0,0)$ is $y - 0 = m_n(x - 0)$,so $y = \left(-\frac{1}{2 - 5h^2}\right)x$. Since $(h, k)$ lies on the normal,we have $k = \left(-\frac{1}{2 - 5h^2}\right)h$,which implies $\frac{k}{h} = \frac{1}{5h^2 - 2}$ (Equation $i$). Since $(h, k)$ lies on the curve,$3k = 6h - 5h^3$,which implies $\frac{k}{h} = \frac{6 - 5h^2}{3}$ (Equation $ii$). Equating $(i)$ and $(ii)$,we get $\frac{1}{5h^2 - 2} = \frac{6 - 5h^2}{3}$. This simplifies to $3 = (6 - 5h^2)(5h^2 - 2) = 30h^2 - 12 - 25h^4 + 10h^2$. Rearranging gives $25h^4 - 40h^2 + 15 = 0$,or $5h^4 - 8h^2 + 3 = 0$. Let $z = h^2$,then $5z^2 - 8z + 3 = 0$. Factoring gives $(5z - 3)(z - 1) = 0$,so $z = \frac{3}{5}$ or $z = 1$. For $z = 1$,$h^2 = 1$,so $h = \pm 1$. The positive integral value is $h = 1$.

If the normal drawn at a point $P$ on the curve $3y = 6x - 5x^3$ passes through $(0,0)$,then the positive integral value of the abscissa of the point $P$ is

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