Q5 Given - Equation is To find- Reduce the above Solution - The standard equation of line So; T T Comparing a = -1 for converting given line into normal form, divide the equation art by + C=o by Ja² +6² - H + २५ √√5 √5 1500 √ G₁1²7621² = √1+4 = √5 ₂x + √5 √5 -2 +24-5 = 0· = √5 and equation to normal form. b=2; c= -5- 17/12 This is normal form of line, P=1 = cos ax+by + c =n. bevey.com/ or W = sin (15)
Q5 Given - Equation is To find- Reduce the above Solution - The standard equation of line So; T T Comparing a = -1 for converting given line into normal form, divide the equation art by + C=o by Ja² +6² - H + २५ √√5 √5 1500 √ G₁1²7621² = √1+4 = √5 ₂x + √5 √5 -2 +24-5 = 0· = √5 and equation to normal form. b=2; c= -5- 17/12 This is normal form of line, P=1 = cos ax+by + c =n. bevey.com/ or W = sin (15)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Q5 Given + Equation is
L
−2+2Y~5 = 0·
To find - Reduce the above equation to normal form.
Solution - The standard equation of line
an+by + C² = n.
Comparing a = 1;
For converting given line into normal form, divide the equation
art by +C =0 by √ a² +6²
√6+1²+(21²
Go;
T
ง
67/20
= √HG = √5
1+ 4
+ २५
√5
ខ
- 1x +
√5
√5
b=2;
1
- 5
and
=
Omal boll
This is normal form of line.
P=1
√√5
cos
√5
or
Sin (2
√5
19verds
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