equation of curve y²-x²+48 +6=0 Let, the equation of Tongent to the curve be y = mx + 1 m where, m = slope of lime Since A (-1,-1) Jdies on this lime, Substitute x=-1₂ cmd y = -1 Im equation (-1) = (-1) = -m - m2 M(-1) + mtm - m² +1 - M-1=0 Solution of Quadrotic equation M 1,2 = 132 = (1 - (-1) ± √ED ²- 4x(+)x) 2x1 1 ± √5 2

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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equation of curve
y²-x+44 +6=0
Let, the equation of Tongent to the curve be
y =
mx+1
m
(11)
where, m = slope of lime
Since A (-1,-1) Jies on this lime,
x=-1₂ cmd y = -1
Im equation
(-1) = M (-1) + 1
(-1)
-m+mm
_m² +1
-m
m² -m-1=0
=
Solution of
M 1,2 =
M1₂2
of Quadrotic equation
=
• (-1) ± √√√ED²- 4x(-1)Xx)
2x1
1 ± √5
2
Substitute
IT
Transcribed Image Text:equation of curve y²-x+44 +6=0 Let, the equation of Tongent to the curve be y = mx+1 m (11) where, m = slope of lime Since A (-1,-1) Jies on this lime, x=-1₂ cmd y = -1 Im equation (-1) = M (-1) + 1 (-1) -m+mm _m² +1 -m m² -m-1=0 = Solution of M 1,2 = M1₂2 of Quadrotic equation = • (-1) ± √√√ED²- 4x(-1)Xx) 2x1 1 ± √5 2 Substitute IT
So, M₁ =
Now
fan d
fand
angle between the Tangent lime (2)
Itama)
11
M₂-
=
LJ
1+Js
=
1-√√5
1+√5 - 1-√s
1 + (1+53) (1-_-5³)
ग्ड
√S
1 + 1-5
M₁-M₂
1 + M₁ M₂
tama = √5 x 4
ग्ड
4+1-5
4√5
2 = tam-¹ (00)
2 = 90°
Soi angle between Two Tangent draw at A (-1,-1) is
90°
Transcribed Image Text:So, M₁ = Now fan d fand angle between the Tangent lime (2) Itama) 11 M₂- = LJ 1+Js = 1-√√5 1+√5 - 1-√s 1 + (1+53) (1-_-5³) ग्ड √S 1 + 1-5 M₁-M₂ 1 + M₁ M₂ tama = √5 x 4 ग्ड 4+1-5 4√5 2 = tam-¹ (00) 2 = 90° Soi angle between Two Tangent draw at A (-1,-1) is 90°
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