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

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(D) Given the curve equation: $ax^2 = 2y^2 - b$. Differentiating both sides with respect to $x$: $2ax = 4y \frac{dy}{dx} \implies \frac{dy}{dx} = \fra

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$2, 0$

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(D) Given the curve equation: $ax^2 = 2y^2 - b$. Differentiating both sides with respect to $x$: $2ax = 4y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2ax}{4y} = \frac{ax}{2y}$. At the point $(1, -1)$,the slope of the tangent is: $m = \frac{a(1)}{2(-1)} = -\frac{a}{2}$. The given line is $x + y = 0$,which can be written as $y = -x$. The slope of this line is $-1$. Since the line touches the curve at $(1, -1)$,their slopes must be equal: $-\frac{a}{2} = -1 \implies a = 2$. Now,substitute $a = 2$ and the point $(1, -1)$ into the original curve equation: $2(1)^2 = 2(-1)^2 - b$ $2 = 2 - b \implies b = 0$. Thus,the values are $a = 2$ and $b = 0$.

If the line $x+y=0$ touches the curve $ax^2 = 2y^2 - b$ at $(1, -1)$,then the values of $a$ and $b$ are respectively:

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