The equation of the normal to the curve
y = (1+x)^2y + cos ²(sin^–1x) at x = 0 is :

Given equation of curve <br><br>y = (1+x)^2y + cos ²(sin^–1x) <br><br>at x = 0 <br><br>$\Rightarrow$ y = (1 + 0)^2y + cos²(sin^–10) <br><br>$\Rightarrow$ y = 1 + 1 <br><br>$\Rightarrow$ y = 2 <br><br>So we have to find the normal at (0, 2) <br><br>Now, y = $${e^{2y\ln \left( {1 + x} \right)}} + {\cos ^2}\left( {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right)$$ <br><br>$\Rightarrow$ y = ${e^{2y\ln \left( {1 + x} \right)}} + {\left( {\sqrt {1 - {x^2}} } \right)^2}$ <br><br>$\Rightarrow$ y = ${e^{2y\ln \left( {1 + x} \right)}} + \left( {1 - {x^2}} \right)$ <br><br>Now differentiate w.r.t. x <br><br>y' = $${e^{2y\ln \left( {1 + x} \right)}}\left[ {2y.\left( {{1 \over {1 + x}}} \right) + \ln \left( {1 + x} \right).2y'} \right]$$ - 2x <br><br>Put x = 0 &amp; y = 2 <br><br>$\Rightarrow$ y' = $${e^{2y\ln \left( {1 + 0} \right)}}\left[ {2y.\left( {{1 \over {1 + 0}}} \right) + \ln \left( {1 + 0} \right).2y'} \right] - 2 \times 0$$ <br><br>$\Rightarrow$ y' = e⁰ [4 + 0] – 0 <br><br>$\Rightarrow$ y' = 4 = slope of tangent to the curve <br><br>so slope of normal to the curve = - ${1 \over 4}$ <br><br>Hence equation of normal at (0, 2) is <br><br>y - 2 = - ${1 \over 4}$(x - 0) <br><br>$\Rightarrow$ 4y – 8 = –x <br><br>$\Rightarrow$ x + 4y = 8