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Question

If $$(x^{2}+y^{2})(a^{2}+b^{2})=(ax+by)^{2}$$ prove that $$\dfrac{x}{a}=\dfrac{y}{b}$$

Solution
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As we know that

$$(a+b)^2=a^2+b^2+2 a b,(a-b)^2=a^2+b^2-2 a b$$

Given $$(x^2+y^2)(a^2+b^2)=(a x+b y)^2$$

$$x^2(a^2+b^2)+y^2(a^2+b^2)=(a x)^2+(b y)^2+2(a x)(b y)$$

$$\implies a^2 x^2+b^2 x^2+a^2 y^2+b^2 y^2=a^2 x^2+b^2 y^2+2 a b x y $$

$$\implies b^2 x^2+a^2 y^2=2 a b x y$$

$$\implies (b x)^2+(a y)^2-2(b x)(a y)=0$$

$$\implies (b x-a y)^2=0$$

$$\implies b x=a y$$

$$\implies \dfrac{x}{a}=\dfrac{y}{b}$$

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