A) \[{{x}^{2}}+{{y}^{2}}2xy=0\]
B) \[{{x}^{2}}+{{y}^{2}}16{{x}^{2}}{{y}^{2}}=0\]
C) \[{{x}^{2}}+{{y}^{2}}4{{x}^{2}}{{y}^{2}}=0\]
D) \[{{x}^{2}}+{{y}^{2}}2{{x}^{2}}{{y}^{2}}=0\]
Correct Answer: C
Solution :
Let the mid point be S(h,k) \[\therefore \]P(2h,0) and Q(0,2k) equation of \[PQ:\frac{x}{2h}+\frac{y}{2k}=1\] \[\because PQ\]is tangent to circle at R(say) \[\therefore \] \[\Rightarrow \frac{1}{4{{h}^{2}}}+\frac{1}{4{{k}^{2}}}=1\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-4{{x}^{2}}{{y}^{2}}=0\] Aliter : tangent to circle \[x\cos \theta +y\sin \theta =1\] \[P:(sec\theta ,0)\] \[Q:(0,cosec\theta )\] \[2h=\sec \theta \Rightarrow \cos \theta =\frac{1}{2h}\And \sin \theta =\frac{1}{2k}\] \[\frac{1}{{{\left( 2x \right)}^{2}}}+\frac{1}{{{\left( 2y \right)}^{2}}}=1\]You need to login to perform this action.
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