A) \[x+y+4=0\]
B) \[x2y+16=0\]
C) \[2xy+2=0\]
D) \[xy+4=0\]
Correct Answer: D
Solution :
tangent to the parabola \[{{y}^{2}}=16x\] is \[y=mx+\frac{4}{m}\] solve it by curve \[xy=-4\] i.e.\[m{{x}^{2}}+\frac{4}{m}x+4=0\] condition of common tangent is D = 0 \[\therefore \]\[{{m}^{3}}=1\]\[\Rightarrow m=1\] \[\therefore \]equation of common tangent is\[~y=x+4\]You need to login to perform this action.
You will be redirected in
3 sec