A) \[2\sqrt{5}\,a\]
B) \[\sqrt{10}\,a\]
C) 10 a
D) \[5\sqrt{2}\,a\]
Correct Answer: D
Solution :
Given equation is \[{{x}^{2}}+{{y}^{2}}+2xy-8ax-8ay-9{{a}^{2}}=0\] or \[{{x}^{2}}+{{y}^{2}}+(-4{{a}^{2}})+2xy\] \[-8ax-8ay-25{{a}^{2}}=0\] or \[{{(x+y-4a)}^{2}}-{{(5a)}^{2}}=0\] or \[(x+y-9a)\,\,(x+y+a)=0\] \[\Rightarrow \] \[x+y-9a=0\]or \[x+y+a=0\] These lines are parallel. Now we find the distance from origin to the line Let \[{{p}_{1}}=\frac{0+0-9a}{\sqrt{{{1}^{2}}+{{1}^{2}}}},\,{{p}_{2}}\frac{0+0+a}{\sqrt{{{1}^{2}}+{{1}^{2}}}}\] \[{{p}_{1}}=-\frac{9a}{\sqrt{2}}\,,\,\,{{p}_{2}}=\frac{a}{\sqrt{2}}\] The distance between two lines is \[|\,{{p}_{2}}-{{p}_{1}}|=\left| \frac{a}{\sqrt{2}}+\frac{9a}{\sqrt{2}} \right|=\frac{10a}{\sqrt{2}}\] \[=5\sqrt{2}\,a\]
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