A) \[{{L}_{1}}>{{L}_{2}}\]
B) \[{{L}_{1}}={{L}_{2}}\]
C) \[{{L}_{1}}<{{L}_{2}}\]
D) \[\frac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{2}\]
Correct Answer: C
Solution :
We have \[{{x}^{2}}+(8x)=9\] \[{{x}^{2}}+9x-x-9=0\] \[x(x+9)-1(x+9)=0\] \[(x+9)(x-1)=0\] \[x=-9,1\]for \[x=1,y=\pm 2\sqrt{2x}=\pm 2\sqrt{2}\] \[{{L}_{1}}=\] Length of AB \[=\sqrt{{{(2\sqrt{2}+2\sqrt{2})}^{2}}+{{(1-1)}^{2}}}=4\sqrt{2}\] \[{{L}_{2}}=\] Length of latus rectum \[=4a=4\times 2=8\]\[{{L}_{1}}<{{L}_{2}}\]You need to login to perform this action.
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