# The equation of the common tangent to the curves y^2 = 16x and xy = - 4 is

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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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verified

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