A) \[3y=9x+2\]
B) \[y=2x+1\]
C) \[2y=x+8\]
D) \[y=x+2\]
Correct Answer: D
Solution :
Any point on \[{{y}^{2}}=8x\] is \[(2{{t}^{2}},\,4t)\] where the tangent is \[yt=x+2{{t}^{2}}.\] Solving it with \[xy=-1,\] \[y(yt-2{{t}^{2}})=-1\] or \[t{{y}^{2}}-2{{t}^{2}}y+1=0\]. For common tangent, it should have equal roots. \ \[4{{t}^{4}}-4t=0\] Þ \[t=0,\,1\]. \[\therefore \] The common tangent is \[y=x+2\], (when \[t=0\], it is \[x=0\] which can touch \[xy=-1\] at infinity only).You need to login to perform this action.
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