A) \[x+2y=42\]
B) \[2x+5y=100\]
C) \[x+3y=58\]
D) \[3x+4y=94\]
Correct Answer: B
Solution :
Let hyperbola \[=\frac{{{x}^{2}}}{36}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\therefore \] (10, 16) lies on it \[\Rightarrow \,\,\,\,\,\,\frac{100}{36}-\frac{256}{{{b}^{2}}}=1\] \[\Rightarrow \,\,\,\,\,\,\frac{64}{36}-\frac{256}{{{b}^{2}}}\] \[\Rightarrow \,\,\,\,\,\,{{b}^{2}}=144\] \[\Rightarrow \,\,\,\,\,\,b=12\] \[\therefore \] Equation of normal \[\frac{x-10}{\frac{10}{36}}=\frac{y-16}{\left( - \right)\frac{16}{144}}\] \[\Rightarrow \] \[2x-20=-5y+80\] \[\Rightarrow \] \[2x+5y=100\]
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