\[\mathbf{13}.\]\[~\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{b}\]

\[\mathbf{x}/{{\mathbf{a}}^{\mathbf{2}~}}+\text{ }\mathbf{y}/{{\mathbf{b}}^{\mathbf{2}~}}=\text{ }\mathbf{2}\]

\[\mathbf{14}.\] \[\mathbf{x}/\mathbf{a}\text{ }=\text{ }\mathbf{y}/\mathbf{b}\]

\[\mathbf{ax}\text{ }+\text{ }\mathbf{by}\text{ }=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}\]

Solution:

Given

\[x/a\text{ }+\text{ }y/b\text{ }\text{ }\left( a\text{ }+\text{ }b \right)~=\text{ }\mathbf{0}\]

\[x/{{a}^{2~}}+\text{ }y/{{b}^{2~}}\text{ }2~=\text{ }\mathbf{0}\]

Cross multiplication Method

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

Comparing the two equations

\[{{a}_{1}}=\frac{1}{a},\]Let \[{{b}_{1}}=\frac{1}{b},\]Let\[{{c}_{1}}=\left( a+b \right)\]

\[{{a}_{2}}=\frac{1}{{{a}^{2}}},{{b}_{2}}=\frac{1}{{{b}^{2}}},{{c}_{2}}=-2\]

$\frac{x}{\frac{-2}{b}+\frac{a}{{{b}^{2}}}+\frac{1}{b}}=\frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}$

\[\frac{x}{\frac{a-b}{{{b}^{2}}}}=\frac{-y}{\frac{-a-b}{{{a}^{2}}}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

\[\frac{x}{\frac{a-b}{{{b}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

$x={{a}^{2}}$

\[\frac{-y}{\frac{-a-b}{{{a}^{2}}}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

$y={{b}^{2}}$

$x={{a}^{2}}$and $y={{b}^{2}}$

Solution:

Given

\[x/a\text{ }\text{ }y/b~=\text{ }\mathbf{0}\]

\[ax\text{ }+\text{ }by\text{ }\text{ }\left( {{a}^{2}}~+\text{ }{{b}^{2}} \right)~=\text{ }\mathbf{0}\]

Cross multiplication Method

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

Comparing the two equations

\[{{a}_{1}}=\frac{1}{a},\]Let \[{{b}_{1}}=\frac{1}{b},\]\[{{c}_{1}}=0\]

\[{{a}_{1}}=a,{{b}_{1}}=b\], Let\[{{c}_{1}}=\left( {{a}^{2}}+{{b}^{2}} \right)\]
$\frac{x}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{-y}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$

$\frac{x}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$

=$x=a$

And,

$\frac{-y}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$

=$y=b$

$x=a$ and $y=b$