Question

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

• A. $x-2y+16=0$
• B. $x+y+4=0$
• C. $2x-y+2=0$
• D. $x-y+4=0$

Answer 1

The correct option is D $x-y+4=0$

Let the equation of the tangent to $y^2=16x$ be
$y=mx+\frac{4}{m}$
This is also a tangent to the curve $xy=-4$
$\Rightarrow x(mx+\frac{4}{m})=-4\Rightarrow m^2{x}^2+4x+4m=0$

For tangent,
$D=0\Rightarrow 16-16m^3=0\Rightarrow m=1$
So, equation of common tangent is $y=x+4$