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Question
Show that the difference between the slopes of the lines given by (tan2θ + cos2θ)x2 - 2xy tan θ + (sin2θ)y2 = 0 is two.
Solution
Comparing the equation
(tan2θ + cos2θ)x2 - 2xy tan θ + (sin2θ)y2 = 0
with ax2 + 2hxy + by2 = 0, we get,
a = tan2θ + cos2θ,
2h = −2tanθ
b = sin2θ
Let m1 and m2 be the slopes of the lines represented by the given equation.
∴ m1 + m2 = `(-"2h")/"b" = - [("- 2 tan" theta)/("sin"^2theta)] = ("2tanθ")/("sin"^2θ)` ....(1)
and m1m2 = `"a"/"b" = ("tan"^2 theta + "cos"^2theta)/"sin"^2theta` .....(2)
∴ (m1 − m2)2 = (m1 + m2)2 −4m1m2 ...[(a − b)2 = (a + b)2 − 4a.b]
`= ((2"tan"theta)/("sin"^2theta))^2 - 4(("tan"^2theta + "cos"^2theta)/("sin"^2theta))`
`= (4 "tan"^2theta)/("sin"^4theta) - 4
(("tan"^2theta + "cos"^2theta)/("sin"^2theta))`
`= (4 (("sin"^2theta)/("cos"^2theta)))/("sin"^4theta) - 4[((("sin"^2theta)/("cos"^2theta) + "cos"^2theta))/("sin"^2theta)]`
`= 4/("sin"^2theta . "cos"^2theta) - 4(("sin"^2theta + "cos"^4theta)/("sin"^2theta . "cos"^2theta))`
`= 4 [(1 - "sin"^2theta - "cos"^4theta)/("sin"^2theta . "cos"^2theta)]`
`= 4 [("cos"^2theta - "cos"^4theta)/("sin"^2theta . "cos"^2theta)]` ...[1+ sin2θ = cos2θ]
`= 4[("cos"^2theta (1 - "cos"^2theta))/("sin"^2theta . "cos"^2theta)]`
= `4[(cos^2theta sin^2theta)/(sin^2theta cos^2theta)]`
∴ (m1 − m2)2 = 4
Taking square root on both sides, we get
∴ |m1 − m2| = 2
∴ the slopes differ by 2.
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