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If θ is the acute angle between the lines given by ax2 + 2hxy + by2 = 0 then prove that tan θ = |2h2-aba+b|. Hence find acute angle between the lines 2x2 + 7xy + 3y2 = 0 - Mathematics and Statistics

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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 

Sum

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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Angle between lines represented by ax2 + 2hxy + by2 = 0
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Chapter 1.4: Pair of Lines - Short Answers II

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