Question

The equation of the common tangent to the curve $y^2=4x$ and $xy=16$ is

• A. $x+4y+16=0$
• B. $x+8y+64=0$
• C. $4x+y+16=0$
• D. $8x+y+64=0$

Answer 1

The correct option is A $x+4y+16=0$

The equation of the tangent to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$
This is also tangent to $xy=16$
$x(mx+\frac{1}{m})=16\Rightarrow m{x}^2+\frac{1}{m}x-16=0△=0\Rightarrow \frac{1}{m^2}+64m=0\Rightarrow 64m^3=-1\Rightarrow m=-\frac{1}{4}$
So,equation of common tangent is $y=-\frac{x}{4}-4\Rightarrow x+4y+16=0$