A) \[\frac{240}{\sqrt{3}}\]
B) \[200\sqrt{3}\]
C) \[240\sqrt{3}\]
D) \[\frac{120}{\sqrt{3}}\]
Correct Answer: C
Solution :
Equation of tangent with slope \[=\frac{-3}{4},\,\] is \[y=\frac{-3}{4}x+C\] Now, \[C=\sqrt{32\times {{\left( \frac{-3}{4} \right)}^{2}}+18}\,=\sqrt{18+18}=6\] (Using condition of tangency) \[\therefore \,\,y=\frac{-3}{4}\,x+6\,\Rightarrow \,3x+4y\,=24\] It meets the coordinate axes in A and B. So A (8, 0) and B(0, 6) Hence, required area \[\Delta AOB=\frac{1}{2}\,(8)\,(6)=24\]You need to login to perform this action.
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