Question

If the equation of common tangent of parabola $y^2=4x$ and circle $x^2+y^2=4$ is $x\pm \lambda y+{\lambda }^2=0$, then the value of $\left[{\lambda }^2\right]$ is

Note: $[.]$ denotes greatest integer function

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

Let equation of the common tangent be $y=mx+c$
This will be a tangent to $x^2+y^2=4$ if
$\left|\frac{c}{\sqrt{1+m^2}}\right|=2\Rightarrow c^2=4(1+m^2)$ ...(1)
The same line will be a tangent to $y^2=4x$ as well, if
$c=\frac{1}{m}$ ...(2)
From $\left(1\right)$ and $\left(2\right)$ we get
$\frac{1}{m^2}=4(1+m^2)\Rightarrow 4m^4+4m^2-1=0$

We get $m^2=\frac{-1+\sqrt{2}}{2}$
As per the question, $x\pm \lambda y+{\lambda }^2=0$ is identical with
$mx-y+\frac{1}{m}=0$
Comparing the coefficients, we get
$\frac{1}{m}=\frac{\pm \lambda }{-1}=m{\lambda }^2$
$\Rightarrow {\lambda }^2=\frac{1}{m^2}=\frac{2}{\sqrt{2}-1}$
$\Rightarrow {\lambda }^2=2\sqrt{2}+2$
$\Rightarrow \left[{\lambda }^2\right]=4$

Hence, option D.