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

Question

The equation of a common tangent to the curves $y^2=16x$ and $xy=-4$, is:

Accepted Answer

Tangent to $y^2=16x$: $y=mx+\dfrac4m$. Substituting into $xy=-4$: $mx^2+\dfrac4mx+4=0$; tangency: $\dfrac{16}{m^2}=16m\Rightarrow m^3=1\Rightarrow m=1$.

$$y=x+4\ \Longrightarrow\ x-y+4=0.$$

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