A) \[{{x}^{2}}-4{{y}^{2}}+16{{x}^{2}}{{y}^{2}}=0\]
B) \[4{{x}^{2}}-{{y}^{2}}+16{{x}^{2}}{{y}^{2}}=0\]
C) \[4{{x}^{2}}-{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\]
D) \[{{x}^{2}}-4{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\]
Correct Answer: D
Solution :
\[4{{y}^{2}}={{x}^{2}}+1\] \[\Rightarrow -{{x}^{2}}+4{{y}^{2}}=1\] \[\Rightarrow -\frac{{{x}^{2}}}{{{1}^{2}}}+\frac{{{y}^{2}}}{{{\left( \frac{1}{2} \right)}^{2}}}=1\] \[a=1,b=\frac{1}{2}\] Let, tangent to the curve is at point \[({{x}_{1}},{{y}_{1}})\]. \[\therefore 4\times 2y\cdot \frac{dy}{dx}=2x\] \[\Rightarrow \frac{dy}{dx}=\frac{2{{x}_{1}}}{8{{y}_{1}}}=\frac{{{x}_{1}}}{4{{y}_{1}}}\] \[\therefore E{{q}^{n}}\] of tangent: \[y=mx+c\] \[\Rightarrow y=\frac{{{x}_{1}}}{4{{y}_{1}}}\cdot x+c\] \[\Rightarrow {{y}_{1}}=\frac{{{x}_{1}}{{y}_{1}}}{4{{y}_{1}}}+c\] \[\Rightarrow c={{y}_{1}}=\frac{x_{1}^{2}}{4{{y}_{1}}}\] \[=\frac{4y_{1}^{2}-x_{1}^{2}}{4{{y}_{1}}}=\frac{1}{4{{y}_{1}}}\] \[\Rightarrow y=\frac{{{x}_{1}}}{4{{y}_{1}}}x+\frac{1}{4{{y}_{1}}}\] \[\Rightarrow 4{{y}_{1}}y={{x}_{x}}x+1......(I)\] Intersects \[x\] axis at \[\left( \frac{-1}{{{x}_{1}}},0 \right)\] And \[y\] axis at \[=\left( 0,\frac{1}{4{{y}_{1}}} \right)\] \[h=\frac{-1}{2{{x}_{1}}}\] \[{{x}_{1}}=\frac{-1}{2h}\] \[{{y}_{1}}=\frac{1}{8k}\] Midpoint: \[\left( \frac{-1}{2{{x}_{1}}},\frac{1}{8{{y}_{1}}} \right):(h,k)\] \[4y_{1}^{2}\ne x_{1}^{2}+1\] \[\Rightarrow 4{{\left( \frac{1}{8k} \right)}^{2}}={{\left( \frac{-1}{2h} \right)}^{2}}+1\] \[\Rightarrow \frac{1}{16{{k}^{2}}}=\frac{1}{4{{h}^{2}}}+1\] \[\Rightarrow 1=\frac{16{{k}^{2}}}{4{{h}^{2}}}+16{{k}^{2}}\] \[{{h}^{2}}=4{{k}^{2}}+16{{h}^{2}}k\] \[{{x}^{2}}-4{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\] This is the required equation.
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