the equation of tangent to the curve y=root x, at a point where tangent drawn to it makes an angle pi/3, with positive x axis?

The given equation of curve is,y = xdydx = 12xslope of the tangent to the curve = tan θ = tan 60° = 3So, 12x = 3x = 112Now, y = x = 123 = 36So, points of intersection are :112 , 36Equation of the tangent to the curve at the point 112 ,36 is,y - 36x - 112 = 36y - 312x - 1 ×126 = 343y - 12x - 1 = 0

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The question is very easy. Look at the information given, i mean tangent makes an angle of 60 degree with positive x axis so it means that at the slope of tangent at the point is tan(60)= root 3

y=root x

on differentiating-

dy/dx=1/(2root x)

dy/dx= slope = tan(60)= root 3

root 3 = 1 / ( 2 root x)

so we get x= 1/12

now you have a value of x. Put this in the equation of cure to get value of y

y= root 1/12 or

y=1/( 2 root 3)

now equation of tangent can be given by point slope form

root 3*[ x- (1/12)]= [y- (1/2 root 3)]

for better understanding please copy and then read

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