The equation of the normal to the curve y=(1+ | vedclass

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(B) Given the equation of the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$.At $x=0$,$y=(1+0)^{2y}+\cos^{2}(\sin^{-1} 0) = 1+1 = 2$.So,we need to find th

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$x+4y=8$

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(B) Given the equation of the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$.At $x=0$,$y=(1+0)^{2y}+\cos^{2}(\sin^{-1} 0) = 1+1 = 2$.So,we need to find the normal at the point $(0, 2)$.Rewrite the equation as $y=e^{2y \ln(1+x)} + (1-x^2)$.Differentiating with respect to $x$:$y' = e^{2y \ln(1+x)} \left[ 2y \cdot \frac{1}{1+x} + \ln(1+x) \cdot 2y' \right] - 2x$.Substituting $x=0$ and $y=2$:$y' = e^{2(2) \ln(1)} \left[ 2(2) \cdot \frac{1}{1+0} + \ln(1) \cdot 2y' \right] - 2(0)$.$y' = e^0 [4 + 0] - 0 = 4$.Thus,the slope of the tangent $m_t = 4$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{4}$.The equation of the normal at $(0, 2)$ is $y - 2 = -\frac{1}{4}(x - 0)$.$4y - 8 = -x$,which simplifies to $x + 4y = 8$.

The equation of the normal to the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$ at $x=0$ is

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