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The equation of the common tangent to the curves y2 = 16x and xy = - 4 is

  • (a) x-y+4=0
  • (b) x+y+4=0
  • (c) x-2y+16=0
  • (d) 2x-y+2=0
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1 Answer

 
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Correct option is (d) y=x+2 

Explaination::

tangents to the curve y2=16x is y=mx+4/m , so it must satisfy xy=-4 

\(x({mx+\frac{4}{m}})=-4\)

\(mx^2+\frac{4}{m}x+4=0 ,\)

since it has equal roots , therefore D=0

\(\frac{16}{m^2}-16m=0\)

\(m^3=1 \)

m=1

therefore , the equation of commom tangent is  y=x+4 i.e. x-y+4=0

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