Question

ii) x2/you+√x2+y2=ay+b√x2+y2
find a and b value
please answer me fast
I will mark him as brilliant
​please

Answer 1

Answer:

For Finding Value of a -

$\frac{x^{2} }{y} + \sqrt{ {x}^{2} } + y^{2} = ay + b \sqrt{x}^{2} + {y}^{2}$

Multiply both sides of equation by y.

${x}^{2} + y \sqrt{x}^{2} + yy^{2} = ayy + b \sqrt{x ^{2} } y + yy^{2}$

To multiply the powers of same of base, add their exponents. Add 1 and 2 to get

3.

$x^{2} + y \sqrt{x}^{2} + y^{3} = ayy^{2} + b \sqrt{x}^{2}y + yy^{2}$

Multiply y and y to get y².

$x^{2} + y \sqrt{x}^{2} + y^{3} = ay^{2} + b \sqrt{x}^{2}y + y {y}^{2}$

$x^{2} + y \sqrt{x}^{2} + y^{3} = ay^{2} + b \sqrt{x ^{2} } y + y {}^{3}$

$ay^{2} = {x}^{2} + y \sqrt{x {}^{2} } + y {}^{3} - b \sqrt{x {}^{2} }y$

$y^{2}a=-by\sqrt{x^{2}}+y\sqrt{x^{2}}+x^{2}$

$\frac{y^{2}a}{y^{2}}=\frac{-by|x|+y|x|+x^{2}}{y^{2}}$

$a=\frac{-by|x|+y|x|+x^{2}}{y^{2}}$

We got value of a .

Now to find value of b.

$\frac{ x ^ { 2 } }{ y } + \sqrt{ x ^ { 2 } } +y ^ { 2 } =ay+b \sqrt{ x ^ { 2 } } +y ^ { 2 }$

$x^{2}+y\sqrt{x^{2}}+yy^{2}=ayy+b\sqrt{x^{2}}y+yy^{2}$

$x^{2}+y\sqrt{x^{2}}+y^{3}=ayy+b\sqrt{x^{2}}y+yy^{2}$

$x^{2}+y\sqrt{x^{2}}+y^{3}=ay^{2}+b\sqrt{x^{2}}y+yy^{2}$

$x^{2}+y\sqrt{x^{2}}+y^{3}=ay^{2}+b\sqrt{x^{2}}y+y^{3}$

$ay^{2}+b\sqrt{x^{2}}y+y^{3}=x^{2}+y\sqrt{x^{2}}+y^{3}$

$b\sqrt{x^{2}}y+y^{3}=x^{2}+y\sqrt{x^{2}}+y^{3}-ay^{2}$

$b\sqrt{x^{2}}y=x^{2}+y\sqrt{x^{2}}+y^{3}-ay^{2}-y^{3}$

$b\sqrt{x^{2}}y=x^{2}+y\sqrt{x^{2}}-ay^{2}$

$y\sqrt{x^{2}}b=-ay^{2}+y\sqrt{x^{2}}+x^{2}$

$\frac{y\sqrt{x^{2}}b}{y\sqrt{x^{2}}}=\frac{-ay^{2}+y|x|+x^{2}}{y\sqrt{x^{2}}}$

$b=\frac{-ay^{2}+y|x|+x^{2}}{y\sqrt{x^{2}}}$

$b=\frac{-ay^{2}+y|x|+x^{2}}{y|x|}$

So , we got value of b.