A) \[{{x}^{2}}+{{y}^{2}}=4{{p}^{2}}\]
B) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{4}{{{p}^{2}}}\]
C) \[{{x}^{2}}+{{y}^{2}}=\frac{4}{{{p}^{2}}}\]
D) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{2}{{{p}^{2}}}\]
Correct Answer: B
Solution :
The straight line \[x\cos \alpha +y\,\sin \alpha =p\] meets the x-axis at the point \[A\left( \frac{p}{\cos \alpha },0 \right)\] and the y-axis at the point \[B\,\left( 0,\,\,\frac{p}{\sin \alpha } \right)\]. Let (h, k) be the coordinates of the middle point of the line segment AB. Then, \[h=\frac{p}{2\,\cos \alpha }\] and \[k=\frac{p}{2\,\sin \alpha }\] \[\Rightarrow \,\,\cos \alpha =\frac{p}{2h}\] and \[\sin \alpha =\frac{p}{2k}\] \[\Rightarrow \,\,{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =\frac{{{p}^{2}}}{4{{h}^{2}}}+\frac{{{p}^{2}}}{4{{k}^{2}}}=1\] Hence locus of the point (h, k) is \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{4}{{{p}^{2}}}\].
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