Let y = m1 x and y = m2 x
be the lines represented by ax2 + 2hxy + by2 = 0. Then,
m₁ + m₂ = -2h/ b and m₁m₂ = a/b … (i)
⇒ tan θ = m₁ - m₂/ (1 + m₁m₂)
⇒ tan θ = √ [(m₁ + m₂)² - 4 m₁m₂]/
1 + m₁m₂
⇒ tan θ = √ [4h²/b²] - [4a/b]/ [1
+ a/b] [Using (i)]
⇒ tan θ = 2√ (h² - ab)/ (a + b)
⇒ θ = tan⁻¹ {2√ (h² - ab)/ (a +
b)}
Thus, the angle θ between the pair of
lines represented by ax2 + 2hxy + by2 = 0 is given by tan
θ = 2√ (h² - ab)/ (a + b)
PROBLEM:
Prove that the angle between the
straight lines given by (x cosα – y sinα)2 = (x2 + y2)
sin2α is 2α.
SOLUTION:
We have,
(x cos α – y sin α)2 = (x2
+ y2) sin2 α
⇒ x2 (cos2 α
– sin2 α) – xy sin 2α + 0.y2 = 0.
Comparing this equation with ax2 +
2hxy + by2 = 0, we get
a= cos2α – sin2 α,
b = 0 and 2h = -sin 2α
Let
be the angle between
the given lines. Then,
⇒ tan θ = 2√ (h² - ab)/ (a + b)
⇒ tan θ = 2√ [(sin²2α)/4 - 0]/
(cos²α - sin²α)
⇒ tan θ = (sin2α)/ (cos2α) = tan2α
⇒ θ = 2α
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