The line x + y = 0 touches the curve y^2 = ax | vedclass

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(C) Step $1$: Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, we have $(-1)^2 = a(1)^3 + b$, which simplifies to $a + b = 1$. Step $2$:

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$a = 2/3, b = 1/3$

Vedclass · Answer

(C) Step $1$: Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, we have $(-1)^2 = a(1)^3 + b$, which simplifies to $a + b = 1$. Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$. Step $3$: At the point $(1, -1)$, the slope of the tangent is $\frac{dy}{dx} = \frac{3a(1)^2}{2(-1)} = -\frac{3a}{2}$. Step $4$: The line $x + y = 0$ can be written as $y = -x$, which has a slope of $-1$. Step $5$: Equating the slopes: $-\frac{3a}{2} = -1$, which gives $a = 2/3$. Step $6$: Substitute $a = 2/3$ into $a + b = 1$: $2/3 + b = 1$, so $b = 1/3$.

The line $x + y = 0$ touches the curve $y^2 = ax^3 + b$ at the point $(1, -1)$. Find the values of $a$ and $b$.

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