The distance of the point (6,-2√2) from the common tangent y=mx+c, m0, of the curves x=2y…

Question

The distance of the point $(6,-2\sqrt2)$ from the common tangent $y=mx+c$, $m>0$, of the curves $x=2y^2$ and $x=1+y^2$ is:

Accepted Answer

$x=2y^2$, i.e. $y^2=\dfrac{x}2$ ($4a=\dfrac12,\ a=\dfrac18$), has tangent $y=mx+\dfrac1{8m}$.

$x=1+y^2$, i.e. $y^2=x-1$ ($4a'=1,\ a'=\dfrac14$), has tangent $y=m(x-1)+\dfrac1{4m}=mx-m+\dfrac1{4m}$.

For a common tangent, the intercepts match: $\dfrac1{8m}=-m+\dfrac1{4m}\Rightarrow\dfrac1{8m}=m\Rightarrow m^2=\dfrac18\Rightarrow m=\dfrac1{2\sqrt2}$ (taking $m>0$).

Then $c=\dfrac1{8m}=\dfrac{\sqrt2}4$, and the line is $x-2\sqrt2y+1=0$.

$$d=\frac{\left|6-2\sqrt2(-2\sqrt2)+1\right|}{\sqrt{1+8}}=\frac{|6+8+1|}{3}=5.$$

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