The angle of intersection of the curves y 2-2 x - and y -sin x isA- -1-B- -1-1 - -C- -11 - -D- -1-

The correct option is A The curves y^2=2xπ and y=sin x intersect at (0,0) and (π 2, 1). Let the gradients of the tangents to the curves be m_1 and m_2 respect ...

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Standard XII

Mathematics

Angle of Intersection of Two Curves

The angle of ...

Question

The angle of intersection of the curves $y^{2} = \frac{2 x}{π} a n d y = s i n x$ is

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Solution

The correct option is B

The curves $y^{2} = \frac{2 x}{π}$ and y=sin x intersect at (0,0) and $\frac{(π}{2, 1)}$ . Let the gradients of the tangents to the curves be $m_{1}$ and $m_{2}$ respectively. Then $m_{1} = \frac{d y}{d x} = \frac{1}{π y} a n d m_{2} = \frac{d y}{d x} = c o s x$

At $(π / 2, 1), m_{1} = \frac{1}{π}$ ,
$m_{2} = c o s \frac{π}{2} = 0$
Thus $t a n θ = \frac{(1 / π) - 0}{1 + (1 / π) (0)} = \frac{1}{π}$

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The angle of intersection of the curves y2=2xπ and y=sinx is

The correct option is B The curves y2=2xπ and y=sin x intersect at (0,0) and (π2,1). Let the gradients of the tangents to the curves be m1 and m2 respectively. Then m1=dydx=1πy and m2=dydx=cosx At(π/2,1),m1=1π, m2=cosπ2=0 Thus tanθ=(1/π)−01+(1/π)(0)=1π

The angle of intersection of the curves $y^{2} = \frac{2 x}{π} a n d y = s i n x$ is