Find the shortest distance between $y^2=4x$ and $y^2=2x-6$ My attempt:-
differentiating $y^2=4x$
$$y_1'=\frac{4}{2y}$$
and doing the same for $y^2=2x-6$
$$y_2'=1/y$$
for the distance to be the least $y'_1=y'_2$ as the distance between  two lines, hence two tangents and thus two curves is minimised when their slopes are the same
which means $y=0$
that happens at the vertex of the parabolas
the distance between the vertices is 3. However, my book says it's $\sqrt5$
where am  I going wrong?
 A: I cannot say where you went wrong, since I do not understand your approach. However, note that if $\left(\frac{x^2}4,x\right)$ is an arbitrary point of the first curve and $\left(\frac{y^2}2+3,y\right)$ is an arbitrary point of the second one, then the square of the distance between them is$$f(x,y)=\left(\frac{x^2}4-\frac{y^2}2-3\right)^2+(x-y)^2.$$But$$\nabla f(x,y)=\left(\frac{x^3}4-2y-x-\frac{xy^2}2,-2x+8y-\frac{x^2y}2+y^3\right)$$and then$$\nabla f(x,y)=(0,0)\iff(x,y)=(0,0)\vee(x,y)=\pm(4,2).$$But $f(0,0)=9>5=f(\pm4,\pm2)$, and therefore the minimum is attained at $\pm(4,2)$. And the minimal distance is $\sqrt{f(\pm4,\pm2)}=\sqrt5$.
A: Here is another way, without requiring the use of $\nabla$.
Following your idea, the tangent to the curve $y^2=4x$ at the point $(t^2, 2t)$ has gradient $\frac1t$.
Set this equal to the gradient of the curve $y^2=2x-6$ at the general point $(x,y)$, which is $\frac1y$, and we have $y=t$.
Therefore the coordinates of the point on the second curve are$$(x,y)=\left(\frac12 t^2+3, t\right)$$
It follows that the distance $h$ between the points on the two curves where the tangents are parallel is given by $$h^2=\left(t^2-\frac12 t^2-3\right)^2+\left(2t-t\right)^2=\left(\frac12t^2-3\right)^2+t^2$$
Differentiating, $$\frac{dh^2}{dt}=2t\left(\frac12t^2-3\right)+2t=0$$
So we have either $t=0\implies h=3$ or $t=2\implies h=\sqrt{5}$
