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

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(C) The slope of the tangent to the curve $y = \frac{x}{x^2 - 3}$ at $(\alpha, \beta)$ is given by $\frac{dy}{dx} = \frac{(x^2 - 3)(1) - x(2x)}{(x^2 -

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

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(C) The slope of the tangent to the curve $y = \frac{x}{x^2 - 3}$ at $(\alpha, \beta)$ is given by $\frac{dy}{dx} = \frac{(x^2 - 3)(1) - x(2x)}{(x^2 - 3)^2} = \frac{-x^2 - 3}{(x^2 - 3)^2}$.At point $(\alpha, \beta)$,the slope is $m = \frac{-\alpha^2 - 3}{(\alpha^2 - 3)^2}$.The given line is $2x + 6y - 11 = 0$,which can be written as $y = -\frac{1}{3}x + \frac{11}{6}$. The slope of this line is $-\frac{1}{3}$.Since the tangent is parallel to the line,$\frac{-\alpha^2 - 3}{(\alpha^2 - 3)^2} = -\frac{1}{3}$.$3(\alpha^2 + 3) = (\alpha^2 - 3)^2 \Rightarrow 3\alpha^2 + 9 = \alpha^4 - 6\alpha^2 + 9 \Rightarrow \alpha^4 - 9\alpha^2 = 0$.Since $\alpha 
eq 0$,we have $\alpha^2 = 9$,so $\alpha = \pm 3$.If $\alpha = 3$,$\beta = \frac{3}{3^2 - 3} = \frac{3}{6} = \frac{1}{2}$.If $\alpha = -3$,$\beta = \frac{-3}{(-3)^2 - 3} = \frac{-3}{6} = -\frac{1}{2}$.For $(\alpha, \beta) = (3, 1/2)$,$|6\alpha + 2\beta| = |6(3) + 2(1/2)| = |18 + 1| = 19$.For $(\alpha, \beta) = (-3, -1/2)$,$|6\alpha + 2\beta| = |6(-3) + 2(-1/2)| = |-18 - 1| = 19$.

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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