For the curve 4x^{5} = 5y^{4},the ratio of th | vedclass

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(C) Given curve: $4x^{5} = 5y^{4}$.Differentiating with respect to $x$: $20x^{4} = 20y^{3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x^{4}}{y^{3}}$

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$\frac{4^{4}}{5^{4}}$

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(C) Given curve: $4x^{5} = 5y^{4}$.Differentiating with respect to $x$: $20x^{4} = 20y^{3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x^{4}}{y^{3}}$.Length of subtangent $(ST) = \left| y \frac{dx}{dy} \right| = \left| y \frac{y^{3}}{x^{4}} \right| = \frac{y^{4}}{x^{4}}$.Length of subnormal $(SN) = \left| y \frac{dy}{dx} \right| = \left| y \frac{x^{4}}{y^{3}} \right| = \frac{x^{4}}{y^{2}}$.We need the ratio of the cube of the subtangent to the square of the subnormal:$\frac{(ST)^{3}}{(SN)^{2}} = \frac{(y^{4}/x^{4})^{3}}{(x^{4}/y^{2})^{2}} = \frac{y^{12}/x^{12}}{x^{8}/y^{4}} = \frac{y^{16}}{x^{20}} = \left(\frac{y^{4}}{x^{5}}\right)^{4}$.Since $4x^{5} = 5y^{4}$,we have $\frac{y^{4}}{x^{5}} = \frac{4}{5}$.Therefore,the ratio is $\left(\frac{4}{5}\right)^{4} = \frac{4^{4}}{5^{4}}$.

For the curve $4x^{5} = 5y^{4}$,the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point is

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