A) 5 units
B) 6 units
C) 8 units
D) 10 units
Correct Answer: A
Solution :
[a] Here, \[9{{x}^{2}}+16{{y}^{2}}=144\] And \[3x+4y=12\] \[\Rightarrow x=\frac{12-4y}{3}\] So, \[9{{\left( \frac{12-4y}{3} \right)}^{2}}+16{{y}^{2}}=144\] On solving we get, \[y=0,3\] For \[y=0;\,\,x=4\] For \[y=3;\,\,x=0\] \[\Rightarrow \] Length of chord \[=\sqrt{{{(0-3)}^{2}}+{{(4-0)}^{2}}}=\sqrt{9+16}\] \[=\sqrt{25}=5\,\,units\]
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