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Question Number 80145 by behi83417@gmail.com last updated on 31/Jan/20

$$\{\begin{array}{l}\frac{\mathbf{x}}{\mathbf{a}}+\frac{\mathbf{y}}{\mathbf{b}}={\mathbf{a}}^2+{\mathbf{b}}^2\\ [\mathbf{a},\mathbf{b}\in \mathbf{R}]\\ \mathbf{ab}({\mathbf{x}}^2-{\mathbf{y}}^2)=\mathbf{xy}({\mathbf{a}}^2-{\mathbf{b}}^2)\end{array}$$

Commented by john santu last updated on 31/Jan/20

$$\frac{x^2-y^2}{xy}=\frac{a^2-b^2}{ab}$$$$\frac{x}{y}-\frac{y}{x}=\frac{a}{b}-\frac{b}{a}$$$$let\frac{x}{y}=t,\frac{a}{b}=p$$$$t-\frac{1}{t}=p-\frac{1}{p}\Rightarrow t-p=\frac{1}{t}-\frac{1}{p}$$$$t-p=-\left(\frac{t-p}{tp}\right)\Rightarrow (t-p)\{1+\frac{1}{tp}\}=0$$$$\left(1\right)t=p\vee tp=-1$$

Commented by john santu last updated on 31/Jan/20

$$\Rightarrow \frac{x}{a}+\frac{y}{b}=a^2+b^2$$$$fort=p\Rightarrow \frac{bx+ay}{ab}=a^2+b^2$$$$\frac{2bx}{ab}=a^2+b^2\Rightarrow x=\frac{a}{2}(a^2+b^2)$$

Commented by john santu last updated on 31/Jan/20

$$ay=bx\Rightarrow y=\frac{b}{a}\times \frac{a}{2}(a^2+b^2)$$$$y=\frac{b}{2}(a^2+b^2)$$$$★donebysantu$$

Commented by jagoll last updated on 31/Jan/20

$$great....mister$$

Commented by behi83417@gmail.com last updated on 31/Jan/20

$$\mathrm{thanks}\mathrm{in}\mathrm{advance}\mathrm{sir}:\mathrm{santu}.$$

Answered by MJS last updated on 31/Jan/20

$$\left(1\right)\Rightarrow y=-\frac{b}{a}x+(a^2+b^2)b$$$$\Rightarrow$$$$\left(2\right)...$$$$x^2-\frac{a(a^2+b^2)(a^2-3b^2)x}{2(a^2-b^2)}-\frac{a^2{(a^2+b^2)}^2{b}^2}{2(a^2-b^2)}=0$$$$\Rightarrow$$$$x=\frac{a(a^2+b^2)}{2}\wedge y=\frac{(a^2+b^2)b}{2}$$$$\vee$$$$x=-\frac{a(a^2+b^2)b^2}{a^2-b^2}\wedge y=\frac{a^2(a^2+b^2)b}{a^2-b^2}$$

Commented by behi83417@gmail.com last updated on 31/Jan/20

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{proph}-:\mathrm{MJS}$$

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