If y = 4x - 6 is a tangent to the curve y^2 = | vedclass

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(D) Given the curve equation: $y^2 = ax^4 + b$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 4ax^3$. Thus,$\frac{dy}{dx} = \fra

Vedclass · Accepted Answer

$a = \frac{4}{9}, b = 0$

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(D) Given the curve equation: $y^2 = ax^4 + b$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 4ax^3$. Thus,$\frac{dy}{dx} = \frac{4ax^3}{2y} = \frac{2ax^3}{y}$. The slope of the tangent at $(3, 6)$ is: $\left. \frac{dy}{dx} \right|_{(3, 6)} = \frac{2a(3)^3}{6} = \frac{2a(27)}{6} = 9a$. The given tangent equation is $y = 4x - 6$,which has a slope of $4$. Equating the slopes: $9a = 4 \Rightarrow a = \frac{4}{9}$. Since the point $(3, 6)$ lies on the curve $y^2 = ax^4 + b$,we substitute the values: $6^2 = a(3)^4 + b \Rightarrow 36 = \frac{4}{9}(81) + b$. $36 = 4(9) + b \Rightarrow 36 = 36 + b \Rightarrow b = 0$. Therefore,$a = \frac{4}{9}$ and $b = 0$.

If $y = 4x - 6$ is a tangent to the curve $y^2 = ax^4 + b$ at $(3, 6)$,then the values of $a$ and $b$ are:

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