A) \[p-\sqrt{{{p}^{2}}+{{q}^{2}}}\]:\[p+\sqrt{{{p}^{2}}+{{q}^{2}}}\]
B) \[p+\sqrt{{{p}^{2}}-{{q}^{2}}}\]:\[p-\sqrt{{{p}^{2}}-{{q}^{2}}}\]
C) \[p:q\]
D) \[p+\sqrt{{{p}^{2}}+{{q}^{2}}}\]:\[p-\sqrt{{{p}^{2}}+{{q}^{2}}}\]
E) \[q+\sqrt{{{p}^{2}}-{{q}^{2}}}\]:\[q-\sqrt{{{p}^{2}}-{{q}^{2}}}\]
Correct Answer: B
Solution :
\[\frac{\frac{x+y}{2}}{\sqrt{xy}}=\frac{p}{q}\] \[\frac{x+y}{2(\sqrt{xy})}=\frac{p}{q}\] ?..(i) \[\frac{{{x}^{2}}+{{y}^{2}}+2xy}{4xy}=\frac{{{p}^{2}}}{{{q}^{2}}}\] \[\frac{{{x}^{2}}+{{y}^{2}}+2xy-4xy}{4xy}=\frac{{{p}^{2}}-{{q}^{2}}}{{{q}^{2}}}\] \[\frac{{{(x-y)}^{2}}}{4xy}=\frac{{{p}^{2}}-{{q}^{2}}}{{{q}^{2}}}\] \[\frac{x-y}{2\sqrt{xy}}=\frac{\sqrt{{{p}^{2}}-{{q}^{2}}}}{q}\] ?..(ii) Equation (i) is divided by (ii), Then \[\frac{x+y}{x-y}=\frac{p}{\sqrt{{{p}^{2}}-{{q}^{2}}}}\]; \[\frac{x}{y}=\frac{p+\sqrt{{{p}^{2}}-{{q}^{2}}}}{p-\sqrt{{{p}^{2}}-{{q}^{2}}}}\].
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