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Question
If θ is the acute angle between the lines given by ax2 + 2hxy + by2 = 0 then prove that tan θ = `|(2sqrt("h"^2) - "ab")/("a" + "b")|`. Hence find acute angle between the lines 2x2 + 7xy + 3y2 = 0
Solution
Let m1 and m2 be the slopes of the lines represented by the equation ax2 + 2hxy + by2 = 0.
∴ m1 + m2 = `(-2"h")/"b"` and m1m2 = `"a"/"b"`
∴ (m1 − m2)2 = (m1 + m2)2 – 4m1m2
= `((-2"h")/"b")^2 - 4("a"/"b")`
= `(4"h"^2)/"b"^2 - (4"ab")/"b"^2`
= `(4"h"^2 - 4"ab")/"b"^2`
= `(4("h"^2 - "ab"))/"b"^2`
∴ m1 − m2 = `± (2sqrt("h"^2 - "ab"))/"b"`
As θ is the acute angle between the lines,
tan θ = `|("m"_1 - "m"_2)/(1 + "m"_1"m"_2)|`
= `|(± (2sqrt("h"^2 - "ab"))/"b")/(1 + "a"/"b")|`
= `|(2sqrt("h"^2 - "ab"))/("a" + "b")|` .......(i)
Comparing the equation 2x2 + 7xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0,
We get a = 2, h = `7/2`, b = 3
Substituting in equation (i), we get
tan θ = `|(2sqrt((7/2)^2 - 2(3)))/(2 + 3)|`
= `|(2sqrt((49/4) - 6))/5|`
= `|(2sqrt((49 - 24)/4))/5|`
= `|(2* 5/2)/5|`
= 1
∴ θ = tan–1(1) = 45°
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