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Mathematics

Find x from the equation :

a+x+a2−x2a+x−a2−x2=bx.\dfrac{a + x + \sqrt{a^2 - x^2}}{a + x - \sqrt{a^2 - x^2}} = \dfrac{b}{x}.

Ratio Proportion

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Answer

Given,

a+x+a2−x2a+x−a2−x2=bx\dfrac{a + x + \sqrt{a^2 - x^2}}{a + x - \sqrt{a^2 - x^2}} = \dfrac{b}{x}

Applying componendo and dividendo,

⇒a+x+a2−x2+a+x−a2−x2a+x+a2−x2−a−x+a2−x2=b+xb−x⇒2(a+x)2a2−x2=b+xb−x⇒(a+x)a2−x2=b+xb−x\Rightarrow \dfrac{a + x + \sqrt{a^2 - x^2} + a + x - \sqrt{a^2 - x^2}}{a + x + \sqrt{a^2 - x^2} - a - x + \sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em] \Rightarrow \dfrac{2(a + x)}{2\sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em] \Rightarrow \dfrac{(a + x)}{\sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em]

Squaring both sides,

⇒(a+x)2a2−x2=(b+x)2(b−x)2⇒(a+x)2(a+x)(a−x)=(b+x)2(b−x)2⇒(a+x)(a−x)=(b+x)2(b−x)2\Rightarrow \dfrac{(a + x)^2}{a^2 - x^2} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \dfrac{(a + x)^2}{(a + x)(a - x)} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \dfrac{(a + x)}{(a - x)} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em]

Again applying componendo and dividendo,

⇒a+x+a−xa+x−a+x=(b+x)2+(b−x)2(b+x)2−(b−x)2⇒2a2x=b2+x2+2bx+b2+x2−2bxb2+x2+2bx−(b2+x2−2bx)⇒2a2x=b2+b2+x2+x2+2bx−2bxb2−b2+x2−x2+2bx−(−2bx)⇒2a2x=2(b2+x2)4bx⇒ax=b2+x22bx\Rightarrow \dfrac{a + x + a - x}{a + x - a + x} = \dfrac{(b + x)^2 + (b - x)^2}{(b + x)^2 - (b - x)^2} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{b^2 + x^2 + 2bx + b^2 + x^2 - 2bx}{b^2 + x^2 + 2bx - (b^2 + x^2 - 2bx)} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{b^2 + b^2 + x^2 + x^2 + 2bx - 2bx}{b^2 - b^2 + x^2 - x^2 + 2bx - (-2bx)} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{2(b^2 + x^2)}{4bx} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{b^2 + x^2}{2bx}

Multiplying both sides by x we get,

⇒a=b2+x22b\Rightarrow a = \dfrac{b^2 + x^2}{2b}

On cross-multiplication,

⇒2ab=b2+x2⇒x2=2ab−b2⇒x=2ab−b2.\Rightarrow 2ab = b^2 + x^2 \\[1em] \Rightarrow x^2 = 2ab - b^2 \\[1em] \Rightarrow x = \sqrt{2ab - b^2}.

Hence, the value of x is 2ab−b2.\sqrt{2ab - b^2}.

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