EQUATION OF TANGENT WHICH IS PARALLEL OR PERPENDICULAR TO THE LINE

Angle between two curves, if they intersect, is defined as the acute angle between the tangent lines to those two curves at the point of intersection.

For the given curves, at the point of intersection using the slopes of the tangents, we can measure

the acute angle between the two curves. Suppose

y = m₁x + c₁ and y = m₂x + c₂

are two lines, then the acute angle θ between these lines is given by,

(i) If the two curves are parallel at (x₁, y₁), then

m₁  =  m₂

(ii) If the two curves are perpendicular at (x₁, y₁) and if m₁ and m₂ exists and finite then

m₁ x m₂  =  -1

Problem 1 :

Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0 .

Solution :

Slope of rectangular hyperbola :

x(dy/dx) + y(1)  =  0

dy/dx  =  -y/x

Slope of parabola :

x² + 4y = 0 

2x+4(dy/dx)  =  0

dy/dx  =  -2x/4

dy/dx  =  -x/2

To find point of intersection of the curves.

y  =  x/2  ----(1) and y  =  -x²/4 ----(2)

-x²/4  =  x/2

-x³  =  8

x³  =  -8

x³  =  (-2)³

x  =  -2

By applying x  =  -2 in (1), we get

y  =  -2/2

y  =  -1

At (-2, -1)

Angle between the above curves  :

m₁ - m₂  =  -1/2 - 1  =  -3/2

m₁ x m₂  =  -1/2 x 1  =  -1/2

tan θ  =  |(-3/2)/1-(-1/2)|

tan θ  =  |(-3/2)/(1/2)|

 θ  =  tan^-1(3)

Problem 2 :

Show that the two curves x² − y² = r² and xy = c² where c, r are constants, cut orthogonally

Solution :

If two two curves are intersecting orthogonally, then 

m₁ x m₂  =  -1

x² − y² = r² 

2x-2y(dy/dx)  =  0

-2y(dy/dx)  =  -2x

m₁  =  dy/dx  =  -x/y  -----(1)

 xy = c²

x(dy/dx) + y(1)  =  0

m₂  =  dy/dx  =  -y/x  -----(2)

m₁ x m₂  =  (-x/y) (-y/x)

=  -1

So, the given curves are intersecting orthogonally.

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