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प्रश्न
Show that the difference between the slopes of the lines given by (tan2θ + cos2θ)x2 − 2xy tan θ + (sin2θ)y2 = 0 is two.
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उत्तर
Comparing the equation,
(tan2θ + cos2θ)x2 − 2xy tan θ + (sin2θ)y2 = 0
with ax2 + 2hxy + by2 = 0, we get,
a = tan2θ + cos2θ, 2h = −2 tanθ, 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]
`= ((2tantheta)/(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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