Question

The shortest distance between the parabolas $2 y^{2}=2 x-1,2 x^{2}=2 y-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
$y=m\left(x-\frac{1}{2}\right)-2\times\frac{1}{4}m-\frac{1}{4}m^{3}$
$\Rightarrow y=mx-\frac{1}{2}m-\frac{1}{2}m-\frac{1}{4}m^{3}$
$\Rightarrow y=mx-m-\frac{m^{3}}{4} \ldots\left(i\right)$
Equation of normal to the curve $x^{2}=y-\frac{1}{2}$ is
$y-\frac{1}{2}=mx+2\left(\frac{1}{4}\right)+\frac{1}{4m^{2}}$
$\Rightarrow y=mx+1+\frac{1}{4m^{2}} \ldots\left(ii\right)$
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 $=\left|\frac{\frac{1}{4}+\frac{1}{4}}{\sqrt{2}}\right|$
$=\frac{1}{2\sqrt{2}}$