Equation of a common tangent to the parabola $y^{2}=4 x$ and the hyperbola $x y=2$ is :

Equation of a common tangent to the parabola $y^{2}=4 x$ and the hyperbola $x y=2$ is :
  1. $x+y+1=0$
  2. $x-2 y+4=0$
  3. $x+2 y+4=0$
  4. $4 x+2 y+1=0$

Solution

Equation of a tangent to parabola $y^{2}=4 x$ is: $y=m x+\frac{1}{m}$ This line is a tangent to $x y=2$ $\therefore \quad x\left(m x+\frac{1}{m}\right)=2 \Rightarrow m x^{2}+\frac{1}{m} x-2=0$ $\because$ Tangent is common for parabola and hyperbola. $\therefore \quad D=\left(\frac{1}{m}\right)^{2}-4 \cdot m \cdot(-2)=0$ $\frac{1}{m^{2}}+8 m=0$ $1+8 m^{3}=0$ $m^{3}=-\frac{1}{8} \Rightarrow m=-\frac{1}{2}$ $\therefore$ Equation of common tangent: $y=-\frac{1}{2} x-2$ $\Rightarrow 2 y=-x-4 \Rightarrow x+2 y+4=0$

Asked in: JEE Main 2019 (11 Jan Shift 1)

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