A) \[{{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=\frac{xy}{6}\]
B) \[{{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=\left( \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{81} \right)\]
C) \[\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)=\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}\]
D) None of the above
Correct Answer: B
Solution :
Let \[(h,\,\,k)\] be the mid-point of a chord of the ellipse\[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\]. Then, its equation is \[\frac{hx}{4}+\frac{ky}{9}-1=\frac{{{h}^{2}}}{9}+\frac{{{k}^{2}}}{9}-1\] \[[\because \,\,T={{S}_{1}}]\] \[\Rightarrow \] \[\frac{hx}{4}+\frac{ky}{9}=\frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9}\] Since, it is a distance of 2 units from the vertex \[(0,\,\,0)\] of the parabola\[{{y}^{2}}=-8ax\]. \[\therefore \] \[\left| \frac{\frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9}}{\sqrt{\frac{{{h}^{2}}}{16}+\frac{{{k}^{2}}}{81}}} \right|=2\] \[\Rightarrow \] \[{{\left( \frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9} \right)}^{2}}=4\left( \frac{{{h}^{2}}}{16}+\frac{{{k}^{2}}}{81} \right)\] Hence, the locus of \[(h,\,\,k)\] is \[{{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=4\left( \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{81} \right)\]
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