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

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

Vedclass · Accepted Answer

$\pm 30$

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(D) Given the curve $2y^3 = ax^2 + x^3$. Differentiating with respect to $x$,we get:$6y^2 \frac{dy}{dx} = 2ax + 3x^2$.At the point $(a, a)$,the slope of the tangent is:$\left( \frac{dy}{dx} ight)_{(a, a)} = \frac{2a(a) + 3(a)^2}{6(a)^2} = \frac{5a^2}{6a^2} = \frac{5}{6}$.The equation of the tangent at $(a, a)$ is:$y - a = \frac{5}{6}(x - a)$$6y - 6a = 5x - 5a$$5x - 6y + a = 0$.To find the intercepts $p$ and $q$ on the coordinate axes,we write the equation in intercept form:$5x - 6y = -a$$\frac{x}{-a/5} + \frac{y}{a/6} = 1$.Thus,$p = -a/5$ and $q = a/6$.Given $p^2 + q^2 = 61$,we have:$\frac{a^2}{25} + \frac{a^2}{36} = 61$$a^2 \left( \frac{36 + 25}{25 	imes 36} ight) = 61$$a^2 \left( \frac{61}{900} ight) = 61$$a^2 = 900$$a = \pm 30$.

If the tangent to the curve $2y^3 = ax^2 + x^3$ at the point $(a, a)$ cuts the coordinate axes at $p$ and $q$ such that $p^2 + q^2 = 61$,then what is the value of $a$?

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