Angle of intersection between two curves is the acute angle between the tangents to the curves at the intersection point. In this article, we discuss how to find the angle of intersection between two curves and the condition for orthogonal curves along with solved examples.
Consider two curves f(x) and g(x). Let (x1, y1) be the point of intersection of these two curves. To find the angle between these two curves, we should draw tangents to these curves at the intersection point. The acute angle between the two tangents is the angle between the given curves f(x) and g(x).
Let m1 be the slope of the tangent to the curve f(x) at (x1, y1).
m1 = df(x)/dx at (x1, y1)
Let m2 be the slope of the tangent to the curve g(x) at (x1, y1).
m2 = dg(x)/dx at (x1, y1)
If θ is the acute angle of intersection between the given curves,
then tan θ = |(m1 – m2)/(1 + m1m2)|
Steps to Find the Angle between two curves
1. Find the point of intersection of the curves by putting the value of y from the first curve into the second curve.
2. Find the slope of tangents, m1 and m2 at the point of intersection.
3. Find the angle between the curves using the formula tan θ = |(m1 – m2)/(1 + m1m2)|
Condition For Orthogonal Curves
1. If m1m2 = -1, then θ = π/2. This means the given curves cut orthogonally at the point (x1, y1) (meet at right angle at the point (x1, y1)).
2. If m1 = 0 and m2 = ∞, then also the curves are orthogonal.
3. If m1 = m2, then the curves touch each other.
Example 1: The angle between the curves xy = 2 and y2 = 4x is
a. tan-1 1/3
b. tan-1 2
c. tan-1 3
d. tan-1 2/3
Given curves xy = 2 ..(i)
y2 = 4x
From (i), y = 2/x
Substitute y = 2/x in (ii)
We get 4/x2 = 4x
=> x3 = 1
=> x = 1
put x = 1 in (i), we get y = 2.
So the point of intersection is (1, 2).
Slope of curve (i), m1 => x dy/dx + y
=> dy/dx = -y/x
m1 = (dy/dx)(1,2) = -2
Slope of curve (ii), m2 => 2y dy/dx = 4
=> dy/dx = 2/y
m2 = (dy/dx)(1,2) = 2/2 = 1
Angle between the given curves, tan θ = |(m1 – m2)/(1 + m1m2)|
=|(-2 – 1)/(1 + -2)|
So θ = tan-1 3.
Hence option c is the answer.
The line tangent to the curves y3-x2y+5y-2x = 0 and x2-x3y2+5x+2y = 0 at the origin intersect at an angle θ equal to
Given curves y3-x2y+5y-2x = 0… (i)
x2-x3y2+5x+2y = 0 …(ii)
Differentiate (i) w.r.t x
3y2 (dy/dx) – 2xy – x2 (dy/dx) + 5 (dy/dx) – 2 = 0
(dy/dx)(3y2 – x2 + 5) = 2 + 2xy
(dy/dx) = (2 + 2xy)/(3y2 – x2 + 5)
m1 = (dy/dx)(0,0) = 2/5
Differentiate (ii) w.r.t x
x2-x3y2+5x+2y = 0
2x-3x2y2-2yx3(dy/dx)+5+2(dy/dx) = 0
=> (dy/dx)(2-2yx3) = (-2x+3x2y2-5)
=> dy/dx = (-2x+3x2y2-5)/(2-2yx3)
m2 = (dy/dx)(0,0) = -5/2
m1m2 = (2/5)(-5/2) = -1
So θ = π/2
Hence option d is the answer.