At which point on the curve 3x^2 - y^2 = 8 is | vedclass

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(B) Given the curve equation: $3x^2 - y^2 = 8$  $(1)$Differentiating with respect to $x$: $6x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x

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$(\pm 2, \pm 2)$

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(B) Given the curve equation: $3x^2 - y^2 = 8$  $(1)$Differentiating with respect to $x$: $6x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x}{y}$.The slope of the tangent at any point $(x, y)$ is $m_t = \frac{3x}{y}$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{y}{3x}$.The given line is $x + 3y = 4$,which can be written as $y = -\frac{1}{3}x + \frac{4}{3}$. The slope of this line is $m = -\frac{1}{3}$.Since the normal is parallel to the line,their slopes must be equal:$-\frac{y}{3x} = -\frac{1}{3} \implies y = x$.Substitute $y = x$ into the curve equation $(1)$:$3x^2 - (x)^2 = 8 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$.Since $y = x$,the points are $(2, 2)$ and $(-2, -2)$.Thus,the points are $(\pm 2, \pm 2)$.

At which point on the curve $3x^2 - y^2 = 8$ is the normal parallel to the line $x + 3y = 4$?

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