The equation of the normal to the curve 3x^2 | vedclass

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Question

(C) Given the curve equation $3x^2 - y^2 = 8$. Differentiating with respect to $x$,we get $6x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \

Vedclass · Accepted Answer

$x + 3y + 8 = 0$

Vedclass · Answer

(C) Given the curve equation $3x^2 - y^2 = 8$. Differentiating with respect to $x$,we get $6x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{3x}{y}$. The slope of the tangent at any point $(x, y)$ is $\frac{3x}{y}$. The slope of the normal is the negative reciprocal of the tangent slope,which is $-\frac{y}{3x}$. The normal is parallel to the line $x + 3y = 10$,which has a slope of $-\frac{1}{3}$. Setting the slopes equal: $-\frac{y}{3x} = -\frac{1}{3} \Rightarrow y = x$. Substituting $y = x$ into the curve equation: $3x^2 - x^2 = 8 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Thus,the points are $(2, 2)$ and $(-2, -2)$. For point $(2, 2)$,the equation of the normal is $y - 2 = -\frac{1}{3}(x - 2) \Rightarrow 3y - 6 = -x + 2 \Rightarrow x + 3y - 8 = 0$. For point $(-2, -2)$,the equation of the normal is $y + 2 = -\frac{1}{3}(x + 2) \Rightarrow 3y + 6 = -x - 2 \Rightarrow x + 3y + 8 = 0$. Comparing with the options,$x + 3y + 8 = 0$ is the correct choice.

The equation of the normal to the curve $3x^2 - y^2 = 8$,which is parallel to the line $x + 3y = 10$,is

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