If the tangent to the curve y = frac{x}{x^2-3 | vedclass

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(C) Given the curve $y = \frac{x}{x^2-3}$.Finding the derivative: $\frac{dy}{dx} = \frac{(x^2-3)(1) - x(2x)}{(x^2-3)^2} = \frac{x^2-3-2x^2}{(x^2-3)^2}

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$|6\alpha + 2\beta| = 19$

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(C) Given the curve $y = \frac{x}{x^2-3}$.Finding the derivative: $\frac{dy}{dx} = \frac{(x^2-3)(1) - x(2x)}{(x^2-3)^2} = \frac{x^2-3-2x^2}{(x^2-3)^2} = \frac{-(x^2+3)}{(x^2-3)^2}$.The slope of the line $2x + 6y - 11 = 0$ is $m = -\frac{2}{6} = -\frac{1}{3}$.Since the tangent is parallel to the line,the slope of the tangent at $(\alpha, \beta)$ is $-\frac{1}{3}$.So,$\frac{-(\alpha^2+3)}{(\alpha^2-3)^2} = -\frac{1}{3} \Rightarrow 3(\alpha^2+3) = (\alpha^2-3)^2$.Let $t = \alpha^2$. Then $3t + 9 = t^2 - 6t + 9 \Rightarrow t^2 - 9t = 0$.Since $(\alpha, \beta) 
eq (0,0)$,$\alpha^2 
eq 0$,so $\alpha^2 = 9$,which means $\alpha = \pm 3$.If $\alpha = 3$,$\beta = \frac{3}{3^2-3} = \frac{3}{6} = \frac{1}{2}$.If $\alpha = -3$,$\beta = \frac{-3}{(-3)^2-3} = \frac{-3}{6} = -\frac{1}{2}$.Now,calculate $|6\alpha + 2\beta|$:For $(\alpha, \beta) = (3, 1/2)$,$|6(3) + 2(1/2)| = |18 + 1| = 19$.For $(\alpha, \beta) = (-3, -1/2)$,$|6(-3) + 2(-1/2)| = |-18 - 1| = |-19| = 19$.Thus,$|6\alpha + 2\beta| = 19$.

If the tangent to the curve $y = \frac{x}{x^2-3}$,$x \in R, (x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0,0)$ on it,is parallel to the line $2x + 6y - 11 = 0$,then

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