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

JEE/Engineering Exams
Maths
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4
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( 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 )
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