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Question
If the angle between the lines represented by ax2 + 2hxy + by2 = 0 is equal to the angle between the lines 2x2 − 5xy + 3y2 = 0, then show that 100(h2 − ab) = (a + b)2
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Solution
Let θ be the acute angle between the lines ax2 + 2hxy + by2 = 0.
tan θ = `|(2sqrt("h"^2 - "ab"))/("a" + "b")|` .......(i)
Comparing the equation 2x2 − 5xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0,
We get a = 2, h = `-5/2`, b = 3
Let α be the acute angle between the lines given by 2x2 − 5xy + 3y2 = 0
∴ tan α = `|(2sqrt((-5/2)^2 - (2)(3)))/(2 + 3)|`
tan α = `|(2sqrt(25/4-6))/(5)|`
= `|(2sqrt((25 - 24)/4))/5|`
= `|(2* 1/2)/5|`
∴ tan α = `1/5` .......(ii)
But θ = α .......[Given]
∴ tan θ = tan α
∴ `|(2sqrt("h"^2 - "ab"))/("a" + "b")| = 1/5` .......[From (i) and (ii)]
By taking square of both sides, we get
`(4("h"^2 - "ab"))/("a" + "b")^2 = 1/25`
∴ 100(h2 – ab) = (a + b)2
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