A) \[x-2y+8=0\]
B) \[x+y-16=0\]
C) \[3x-y-16=0\]
D) \[x-3y+16=0\]
Correct Answer: A
Solution :
\[\left. \begin{matrix} x-y+2=0 \\ 2x-y-2=0 \\ \end{matrix} \right\}\Rightarrow (4,6)\] St. line passing through (4, 6) is \[(y-6)=m(x-4)\] \[y=mx+6-4m,\] which is tangent to the parabola\[{{y}^{2}}=8x\]. \[\therefore 6-4m=\frac{2}{m}\Rightarrow 3m-2{{m}^{2}}=1\] \[\Rightarrow 2{{m}^{2}}-3m+1=0\Rightarrow m=\frac{1}{2},1\] \[\therefore \]\[y=\frac{1}{2}x+6-2\Rightarrow 2y=x+8\] \[x-2y+8=0\]You need to login to perform this action.
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