Find the acute angle between the lines
4x^2+ 5xy + 4y^2=0
Answers
Answered by
14
Step-by-step explanation:
Given Find the acute angle between the lines 4x^2+ 5xy + 4y^2=0
- Now we need to find the acute angle between the lines 4x^2 + 5xy + 4y^2 = 0
- Now this is in the form of ax^2 + 2hxy + by^2 = 0
- Now for angle we have tan theta = l2 √h^2 – ab / a + b l
- So a = 4, 2h = 5 or h = 5/2, b = 4
- Now tan theta = l 2 √(5/2)^2 - 16 / 4 + 4 l
- = l 2 √ 25 / 4 – 16 / 8 l
- = l 2 √39 / 4 / 8
- = √39 / 8
Therefore tan theta = √39 / 8
Reference link will be
https://brainly.in/question/16422627
Answered by
2
Comparing the equation
4x
2
+5xy+y
2
=0 with
ax
2
+2hxy+by
2
=0, we get,
a=4,2h=5, i.e.h=
2
5
and b=1.
Let θ be the acute angle between the lines.
∴tanθ=
∣
∣
∣
∣
∣
∣
a+b
2
h
2
−ab
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
4+1
2
(
2
5
)
2
−4(1)
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
5
2
(
4
25
)−4
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
5
2×
2
3
∣
∣
∣
∣
∣
∣
∣
∣
∴tanθ=
5
3
∴θ=tan
−1
(
5
3
)
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