If the tangent to the curve 2y^3 = ax^2 + x^3 | vedclass

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(B) Given curve: $2y^3 = ax^2 + x^3$ $(i)$ Point of tangency: $(a, a)$ Differentiating $(i)$ with respect to $x$: $6y^2 \frac{dy}{dx} = 2ax + 3x^2$ At

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$30$

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(B) Given curve: $2y^3 = ax^2 + x^3$ $(i)$ Point of tangency: $(a, a)$ Differentiating $(i)$ with respect to $x$: $6y^2 \frac{dy}{dx} = 2ax + 3x^2$ At point $(a, a)$: $6a^2 \frac{dy}{dx} = 2a^2 + 3a^2 = 5a^2$ $\frac{dy}{dx} = \frac{5a^2}{6a^2} = \frac{5}{6}$ Equation of the tangent line at $(a, a)$ with slope $m = \frac{5}{6}$: $y - a = \frac{5}{6}(x - a)$ $6y - 6a = 5x - 5a$ $5x - 6y = -a$ Dividing by $-a$: $\frac{x}{-a/5} + \frac{y}{a/6} = 1$ Comparing with intercept form $\frac{x}{\alpha} + \frac{y}{\beta} = 1$,we get $\alpha = -\frac{a}{5}$ and $\beta = \frac{a}{6}$. Given $\alpha^2 + \beta^2 = 61$: $(-\frac{a}{5})^2 + (\frac{a}{6})^2 = 61$ $\frac{a^2}{25} + \frac{a^2}{36} = 61$ $\frac{36a^2 + 25a^2}{900} = 61$ $\frac{61a^2}{900} = 61$ $a^2 = 900$ $|a| = 30$.

If the tangent to the curve $2y^3 = ax^2 + x^3$ at the point $(a, a)$ cuts off intercepts $\alpha$ and $\beta$ on the coordinate axes,where $\alpha^2 + \beta^2 = 61$,then the value of $|a|$ is

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