A) \[y-3x+4=0\]
B) \[3y-x+36=0\]
C) \[3y+x-36=0\]
D) \[3y+x+36=0\]
Correct Answer: D
Solution :
Line perpendicular to given line, \[3y+x=\lambda \] \[y=\frac{-1}{3}x+\frac{\lambda }{3}\]. Here, \[m=\frac{-1}{3},\,\,c=\frac{\lambda }{3}\] If we compare \[{{y}^{2}}=16x\] with \[{{y}^{2}}=4ax\] then \[a=4\], Condition for tangency is, \[c=\frac{a}{m}\Rightarrow \frac{\lambda }{3}=\frac{4}{(-1/3)}\Rightarrow \lambda =-36\] Required equation is; \[x+3y+36=0\].
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