a/x-b/y=0 ab2/x+a2b/y=a2+b2 by cross multiplication

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Lipakshi Kerketta 6 years, 6 months ago
- 1 answers
Related Questions
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Sia ? 6 years, 6 months ago
The given equations are
{tex}\frac { a } { x } - \frac { b } { y } = 0{/tex} ........... (i)
{tex}\frac { a b ^ { 2 } } { x } + \frac { a ^ { 2 } b } { y } = \left( a ^ { 2 } + b ^ { 2 } \right){/tex} ........... (ii)
Put {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex}= v in the equation (i) and (ii), we get
{tex}\Rightarrow a u - b v = 0{/tex} .............(iii)
{tex}ab^2u + a^2bv - (a^2 + b^2) = 0{/tex}......(iv)
So, by cross multiplication,
{tex}\frac { u } { b _ { 1 } c _ { 2 } - b _ { 2 } c _ { 1 } } = \frac { v } { c _ { 1 } a _ { 2 } - c _ { 2 } a _ { 1 } } = \frac { 1 } { a _ { 1 } b _ { 2 } - a _ { 2 } b _ { 1 } }{/tex}
{tex}\Rightarrow \frac {u } { ( - b ) \left[ - \left( a ^ { 2 } + b ^ { 2 } \right) \right] - \left( a ^ { 2 } b \right) ( 0 ) }{/tex}{tex}= \frac { v } { ( 0 ) \left( a ^ { 2 } b \right) - \left[ - \left( a ^ { 2 } + b ^ { 2 } \right) \right] ( a ) } = \frac { 1 } { ( a ) \left( a ^ { 2 } b \right) - \left( a ^ { 2 } b \right) ( - b ) }{/tex}
{tex}\Rightarrow \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { v } { a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex} \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex}\Rightarrow \mathrm { u } = \frac { 1 } { \mathrm { a } }{/tex}
and {tex} \frac { v } { a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex}\Rightarrow \mathrm { v } = \frac { 1 } { \mathrm { b } }{/tex}
If {tex} \mathrm { u } = \frac { 1 } { \mathrm { a } }{/tex}
{tex}\Rightarrow{/tex} {tex}x = a{/tex}
{tex} \mathrm { v } = \frac { 1 } { \mathrm { b } }{/tex}
{tex}\Rightarrow{/tex} {tex}y = b{/tex}
0Thank You