Question

The shortest distance between the parabolas $y^2=4x$ and $y^2=2x-6$ is

• A. $2$
• B. $\sqrt{5}$
• C. $3$
• D. $noneofthese$

Answer 1

Answer: (B)

$\sqrt{5}$

Since the shortest distance between the two curves happens to be at the normal which is common to both the cuves.

Therefore

The normal to the curve $y^2=4x$ at $(m^2,2m)$ is given by:

$(y-2m)=-m(x-m^2)$

i.e., $y-2m=-mx+m^3$

i.e., $y-2m+mx-m^3=0$

i.e., $y+mx-2m-m^3=0$

And the normal to the curve $y^2=2x-6$ at $(\frac{1}{2}{m}^2+3,m)$ is given by:

$(y-m)=-m(x-\frac{1}{2}{m}^2-3)$

i.e., $y-m=-mx+\frac{1}{2}{m}^3+3m$

i.e., $y-m+mx-\frac{1}{2}{m}^3-3m=0$

i.e., $y+mx-4m-\frac{1}{2}{m}^3=0$

These two normals are common if:

$y+mx-2m-m^3=$ $y+mx-4m-\frac{1}{2}{m}^3$

i.e., $-2m-m^3=$ $-4m-\frac{1}{2}{m}^3$

i.e., $m^3-4m=0$

i.e., $m(m^2-4)=$

i.e., $m(m+2)(m-2)=$

Therefore, $m=0,2,-2$

Thus, the points are: $(4,4)$ and $(5,2)$

And the distance is: $d=\sqrt{(5-4{)}^2+(2-4{)}^2}$

i.e., $d=\sqrt{1+4}$

i.e., $d=\sqrt{5}$