Question

The shortest distance between the parabolas $2y^2=2x-1,2x^2=2y-1$ is

• A. $2\sqrt{2}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $4$
• D. $\sqrt{\frac{36}{5}}$

Answer 1

Answer: (B)

We have,
$2y^2=2x-1$ and $2x^2=2y-1$
$\Rightarrow y^2=x-\frac{1}{2}$ and $x^2=y-\frac{1}{2}$
Now, the shortest distance always along common normal to curve
Equation of normal to the curve $y^2=x-\frac{1}{2}$ is
$y=m\left(x-\frac{1}{2}\right)-2\times \frac{1}{4}m-\frac{1}{4}{m}^3$
$\Rightarrow y=mx-\frac{1}{2}m-\frac{1}{2}m-\frac{1}{4}{m}^3$
$\Rightarrow y=mx-m-\frac{m^3}{4}\dots \left(i\right)$
Equation of normal to the curve $x^2=y-\frac{1}{2}$ is
$y-\frac{1}{2}=mx+2\left(\frac{1}{4}\right)+\frac{1}{4m^2}$
$\Rightarrow y=mx+1+\frac{1}{4m^2}\dots \left(ii\right)$
Since, Eqs. $\left(i\right)$ and $\left(ii\right)$ are same normal.
$\therefore -m-\frac{m^3}{4}=1+\frac{1}{4m^2}$
$\Rightarrow \frac{-4m-m^3}{4}=\frac{4m^2+1}{4m^2}$
$\Rightarrow m^5+4m^3+4m^2+1=0$
$\Rightarrow (m+1)(m^4-m^3+5m^2-m+1)=0$
$\Rightarrow m=-1$
Hence, slope of tangent $=1$
Equation of tangent to curve $y^2=x-\frac{1}{2}$ is
$y=\left(x-\frac{1}{2}\right)+\frac{1}{4}$
$\Rightarrow y=x-\frac{1}{4}$ Equation of tangent to curve $x^2=y-\frac{1}{2}$ is
$y=x+\frac{1}{4}$
$\therefore$ Required distance $=\left|\frac{\frac{1}{4}+\frac{1}{4}}{\sqrt{2}}\right|$
$=\frac{1}{2\sqrt{2}}$