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

Question

The equation of the normal to the curve

y = (1+x)2y + cos
2(sin--1x) at x = 0 is :

Accepted Answer

Given equation of curve

y = (1+x)2y + cos
2(sin--1x)

at x = 0

$$ \Rightarrow $$ y = (1 + 0)2y + cos2(sin--10)

$$ \Rightarrow $$ y = 1 + 1

$$ \Rightarrow $$ y = 2

So we have to find the normal at (0, 2)

Now, y = $${e^{2y\ln \left( {1 + x} \right)}} + {\cos ^2}\left( {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right)$$

$$ \Rightarrow $$ y = $${e^{2y\ln \left( {1 + x} \right)}} + {\left( {\sqrt {1 - {x^2}} } \right)^2}$$

$$ \Rightarrow $$ y = $${e^{2y\ln \left( {1 + x} \right)}} + \left( {1 - {x^2}} \right)$$

Now differentiate w.r.t. x

y' = $${e^{2y\ln \left( {1 + x} \right)}}\left[ {2y.\left( {{1 \over {1 + x}}} \right) + \ln \left( {1 + x} \right).2y'} \right]$$ - 2x

Put x = 0 \& y = 2

$$ \Rightarrow $$ y' = $${e^{2y\ln \left( {1 + 0} \right)}}\left[ {2y.\left( {{1 \over {1 + 0}}} \right) + \ln \left( {1 + 0} \right).2y'} \right] - 2 \times 0$$

$$ \Rightarrow $$ y' = e0
[4 + 0] -- 0

$$ \Rightarrow $$ y' = 4 = slope of tangent to the curve

so slope of normal to the curve = - $${1 \over 4}$$

Hence equation of normal at (0, 2) is

y - 2 = - $${1 \over 4}(x - 0)

\[ \Rightarrow $$ 4y -- 8 = --x

$$ \Rightarrow $$ x + 4y = 8

Solutions · 1