question_answer

Let \[{{L}_{1}}\]be the length of the common chord of the curves \[{{x}^{2}}+{{y}^{2}}=9\] and\[{{y}^{2}}=8x,\] and \[{{L}_{2}}\] be the length of the latus rectum of \[{{y}^{2}}=8x,\] then:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

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 \[=4a=4\times 2=8\]\[{{L}_{1}}<{{L}_{2}}\]