A) \[\frac{2}{17}\]
B) \[\frac{1}{17}\]
C) \[\frac{8}{17}\]
D) \[\frac{4}{17}\]
Correct Answer: A
Solution :
| \[\frac{{{x}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}=1\] |
| Let \[(a,b)\] is \[(\sqrt{2}cos\theta ,2\sqrt{2}sin\theta )\] tangent at\[(1,2)\]is \[4x+2y=8\]\[\Rightarrow \] \[2x+y=4\]\[\Rightarrow \] slope is\[-2={{m}_{1}}\] |
| tangent at\[\left( \sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta \right)\]is\[4\sqrt{2}\cos \theta x+2\sqrt{2}\sin \theta y=8\]\[\Rightarrow \]slope is\[-2\cot \theta ={{m}_{2}}\] |
| Now, \[{{m}_{1}}{{m}_{2}}=-1\]\[\Rightarrow \]\[4\cot \theta =-1\]\[\Rightarrow \]\[cos\theta =\frac{1}{\sqrt{17}}or\frac{-1}{\sqrt{17}}\]\[\Rightarrow \]\[a=\sqrt{\frac{2}{17}}or-\sqrt{\frac{2}{17}}\]\[\Rightarrow \]\[{{a}^{2}}=\frac{2}{17}.\] |
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