A) \[2x-3y=1\]
B) \[x=0\]
C) \[x=1\]
D) \[y=0\]
Correct Answer: D
Solution :
We know that, the equation of normal at the point \[({{x}_{1}},{{y}_{1}})\] to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[\frac{{{a}^{2}}x}{{{x}_{1}}}+\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}+{{b}^{2}}\] Given equation is \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\] Here, \[{{a}^{2}}=16,{{b}^{2}}=9\] \[\therefore \] The equation of normal at the point \[(-4,0)\]is \[\frac{16x}{-4}+\frac{9y}{0}=16+9\] \[\Rightarrow \] \[\frac{9y}{0}=25+\frac{16x}{4}\] \[\Rightarrow \] \[9y=0\Rightarrow y=0\]
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