If all the normals drawn to the curve y=frac{ | vedclass

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(A) Given the curve $y=\frac{1+3x^2}{3+x^2}$.To find the points of intersection with $y=1$,we solve:$\frac{1+3x^2}{3+x^2} = 1$ $\Rightarrow 1+3x^2 = 3

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$4$

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(A) Given the curve $y=\frac{1+3x^2}{3+x^2}$.To find the points of intersection with $y=1$,we solve:$\frac{1+3x^2}{3+x^2} = 1$ $\Rightarrow 1+3x^2 = 3+x^2$ $\Rightarrow 2x^2 = 2$ $\Rightarrow x = \pm 1$.Thus,the points of intersection are $(1, 1)$ and $(-1, 1)$.Now,find the derivative: $\frac{dy}{dx} = \frac{(3+x^2)(6x) - (1+3x^2)(2x)}{(3+x^2)^2} = \frac{18x + 6x^3 - 2x - 6x^3}{(3+x^2)^2} = \frac{16x}{(3+x^2)^2}$.At $x=1$,$\frac{dy}{dx} = \frac{16}{16} = 1$. The slope of the normal is $m_1 = -1$.The equation of the normal at $(1, 1)$ is $y-1 = -1(x-1) \Rightarrow y+x = 2$.At $x=-1$,$\frac{dy}{dx} = \frac{-16}{16} = -1$. The slope of the normal is $m_2 = -(-1)^{-1} = 1$.The equation of the normal at $(-1, 1)$ is $y-1 = 1(x+1) \Rightarrow y-x = 2$.Solving $y+x=2$ and $y-x=2$,we get $2y=4 \Rightarrow y=2$ and $x=0$.Thus,$(\alpha, \beta) = (0, 2)$.Therefore,$3\alpha+2\beta = 3(0) + 2(2) = 4$.

If all the normals drawn to the curve $y=\frac{1+3x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha, \beta)$,then $3\alpha+2\beta=$

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