A) \[y=0\]
B) \[y=x\]
C) \[x=0\]
D) \[x=-y\]
Correct Answer: A
Solution :
\[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\]\[\Rightarrow \]\[\frac{2x}{16}-\frac{2y}{9}\frac{dy}{dx}=0\] Þ \[\frac{dy}{dx}=\frac{2x\times 9}{16\times 2y}\]\[=\frac{9}{16}\frac{x}{y}\] Þ \[{{\left( \frac{-dx}{dy} \right)}_{(-4,0)}}=\frac{-16}{9}\frac{y}{x}=0\] Hence, equation of normal Þ \[(y-0)\,=0(x+4)\]Þ\[y=0.\]You need to login to perform this action.
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