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प्रश्न
Find the measure of the acute angle between the line represented by:
(a2 - 3b2)x2 + 8abxy + (b2 - 3a2)y2 = 0
उत्तर
Comparing the equation
(a2 - 3b2)x2 + 8abxy + (b2 - 3a2)y2 = 0 with
Ax2 + 2Hxy + by2 = 0, we have,
A = a2 - 3b2, 2H = 8ab and B = b2 - 3a2
∴ H2 - AB = (4ab)2 - (a2 - 3b2) (b2 - 3a2)
= 16a2b2 - (a2b2 - 3a4 - 3b4 + 9a2b2)
= 16a2b2 - 10a2b2 + 3a4 + 3b4
= 3a4 + 6a2b2 + 3b4
= 3(a4 + 2a2b2 + b4)
= 3(a2 + b2)2
∴ `sqrt("H"^2 - "AB") = sqrt3 ("a"^2 + "b"^2)`
Also, A + B = a2 - 3b2 + b2 - 3a2
= - 2a2 - 2b2
= - 2(a2 + b2)
Let θ be the acute angle between the lines.
∴ tan θ = `|(2 sqrt("H"^2 - "AB"))/("A + B")|`
`= |(2sqrt3("a"^2 + "b"^2))/(- 2("a"^2 + "b"^2))|`
= `sqrt3` = tan 60°
∴ θ = 60°
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