Question

( y=-int x+2 )
( frac{d y}{d x}=-1 )
( 2 sqrt{x} )
Now, on puting ( y=x )
( x=-frac{x+2}{x+2} )( quad y=tan left(t a n^{-1} xright) )
( x in R )
( y=x ) t ( x in R )
( x-2=-sqrt{x} )
( x^{2}+4-4 x=x )
( x^{2}-5 x+4=0 )
( x^{2}-4 x-x+4=0 )
( x(x-4)-1(x-4)=0 )
( (x-1)(x-4=0 )
( x=1,4 )
( left.frac{d y}{d x}right|_{x=1}=frac{-1}{2 sqrt{1}}=-frac{1}{2} )
( left.frac{d y}{d x}right|_{x=4}=frac{-1}{2 sqrt{4}}=-frac{1}{4} )
( begin{aligned} text { whe know }, & begin{array}{r}m, x m_{2}=-1,-frac{1}{4} x m=-1 -frac{1}{2} times m_{2}=-1 Rightarrow & =frac{m_{2}=2}{sqrt{m},=4}end{array} end{aligned} )
( y-1=2(x-1) )
( y-1=2 x-2 )
(1) ( m=2, a+(1,1) )
(ii) ( m=4 quad a(4,4) )
( y-y=4(x-9) )
( y-4=4 x-16 )
( 2 x-9 y-1=0 quad y-y-y-12=0 )

# Equation of the normal to the curve y=-(x +2 at the point of its intersection with the curve y = tan (tan'x) is

Solution