If the tangent of the curve 4y^3 = 3ax^2 + x^ | vedclass

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(B) Given the curve $4y^3 = 3ax^2 + x^3$. Differentiating with respect to $x$,we get $12y^2 \frac{dy}{dx} = 6ax + 3x^2$. At the point $(a, a)$,the slo

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$\pm 5$

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(B) Given the curve $4y^3 = 3ax^2 + x^3$. Differentiating with respect to $x$,we get $12y^2 \frac{dy}{dx} = 6ax + 3x^2$. At the point $(a, a)$,the slope $m = \frac{dy}{dx} = \frac{6a(a) + 3a^2}{12a^2} = \frac{9a^2}{12a^2} = \frac{3}{4}$. The equation of the tangent at $(a, a)$ is $y - a = \frac{3}{4}(x - a)$,which simplifies to $4y - 4a = 3x - 3a$,or $3x - 4y + a = 0$. The intercepts on the coordinate axes are $x = -\frac{a}{3}$ and $y = \frac{a}{4}$. The area of the triangle formed with the axes is $\frac{1}{2} |x_{int} 	imes y_{int}| = \frac{1}{2} |(-\frac{a}{3}) 	imes (\frac{a}{4})| = \frac{a^2}{24}$. Given the area is $\frac{25}{24}$,we have $\frac{a^2}{24} = \frac{25}{24}$,which implies $a^2 = 25$,so $a = \pm 5$.

If the tangent of the curve $4y^3 = 3ax^2 + x^3$ drawn at the point $(a, a)$ forms a triangle of area $\frac{25}{24}$ sq. units with the coordinate axes,then $a =$

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