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(C) The given equation of the curve is $x^2(x - y) + a^2(x + y) = 0$,which can be expanded as $x^3 - x^2y + a^2x + a^2y = 0$.To find the tangent at th

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$x + y = 0$

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(C) The given equation of the curve is $x^2(x - y) + a^2(x + y) = 0$,which can be expanded as $x^3 - x^2y + a^2x + a^2y = 0$.To find the tangent at the origin $(0, 0)$,we equate the lowest degree terms to zero.The lowest degree terms in the equation are $a^2x + a^2y = 0$.Dividing by $a^2$ (assuming $a 
eq 0$),we get $x + y = 0$.Alternatively,differentiating with respect to $x$:$3x^2 - (2xy + x^2 \frac{dy}{dx}) + a^2 + a^2 \frac{dy}{dx} = 0$.At the origin $(0, 0)$,this becomes $0 - 0 + a^2 + a^2 \frac{dy}{dx} = 0$.$a^2 \frac{dy}{dx} = -a^2$,so $\frac{dy}{dx} = -1$.The equation of the tangent is $y - 0 = -1(x - 0)$,which simplifies to $y = -x$ or $x + y = 0$.

What is the equation of the tangent to the curve $x^2(x - y) + a^2(x + y) = 0$ at the origin?

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