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

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(C) Differentiate $2y^3 = x^3 + ax^2$ with respect to $x$: $6y^2 \frac{dy}{dx} = 3x^2 + 2ax$.At the point $(a, a)$, the slope $m = \frac{dy}{dx} = \fr

Vedclass · Accepted Answer

$\pm 30$

Vedclass · Answer

(C) Differentiate $2y^3 = x^3 + ax^2$ with respect to $x$: $6y^2 \frac{dy}{dx} = 3x^2 + 2ax$.At the point $(a, a)$, the slope $m = \frac{dy}{dx} = \frac{3a^2 + 2a^2}{6a^2} = \frac{5a^2}{6a^2} = \frac{5}{6}$.The equation of the tangent at $(a, a)$ is $y - a = \frac{5}{6}(x - a)$.Multiplying by $6$, we get $6y - 6a = 5x - 5a$, which simplifies to $5x - 6y = -a$.To find the intercepts, rewrite the equation as $\frac{x}{-a/5} + \frac{y}{a/6} = 1$.Thus, the intercepts are $\alpha = -a/5$ and $\beta = a/6$.Given $\alpha^2 + \beta^2 = 61$, we have $\frac{a^2}{25} + \frac{a^2}{36} = 61$.$\frac{36a^2 + 25a^2}{900} = 61 \implies \frac{61a^2}{900} = 61$.$a^2 = 900 \implies a = \pm 30$.

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

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