If a variable tangent to the curve x^2y = c^3 | vedclass

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(C) Given the curve equation $x^2y = c^3$.Differentiating with respect to $x$,we get $x^2 \frac{dy}{dx} + 2xy = 0$,which implies $\frac{dy}{dx} = -\fr

Vedclass · Accepted Answer

$\frac{27}{4}c^3$

Vedclass · Answer

(C) Given the curve equation $x^2y = c^3$.Differentiating with respect to $x$,we get $x^2 \frac{dy}{dx} + 2xy = 0$,which implies $\frac{dy}{dx} = -\frac{2y}{x}$.The equation of the tangent at point $(x, y)$ is $Y - y = -\frac{2y}{x}(X - x)$.To find the $x$-intercept $a$,set $Y = 0$: $-y = -\frac{2y}{x}(a - x) \implies x = 2(a - x) \implies 3x = 2a \implies a = \frac{3x}{2}$.To find the $y$-intercept $b$,set $X = 0$: $b - y = -\frac{2y}{x}(0 - x) \implies b - y = 2y \implies b = 3y$.Now,calculate $a^2b$: $a^2b = (\frac{3x}{2})^2(3y) = \frac{9x^2}{4} \cdot 3y = \frac{27}{4}x^2y$.Since $x^2y = c^3$,we have $a^2b = \frac{27}{4}c^3$.

If a variable tangent to the curve $x^2y = c^3$ makes intercepts $a$ and $b$ on the $x$-axis and $y$-axis respectively,then the value of $a^2b$ is

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