Minimum distance

The shortest distance the straight line and the parabola is between the straight line and a point on the parabola (the upper half) where the tangent at that point is parallel to the line (see figure).

The gradient of the straight line is given by $1 - 2\dfrac {dy}{dx} = 0 \implies \dfrac {dy}{dx} = \dfrac 12$ . The gradient of the parabola is given by $2 y \dfrac {dy}{dx} = 4 \implies \dfrac {dy}{dx} = \dfrac 2y$ . The point on the parabola having a gradient of the straight line is given by $\dfrac 2y = \dfrac 12 \implies y = 4$ and $x=4$ . The shortest distance is on the straight line perpendicular to the tangent at the point whose equation is $\dfrac {y-4}{x-4} = -\blue {-2}$ , where $-\blue {-2}$ is the gradient of the perpendicular straight line; $\implies y = 12-2x$ . The perpendicular line intersects the straight line at $x-24+4x + 10 = 0$ , $\implies x = 2.8$ and $y = 6.4$ . And the shortest distance is $\sqrt{(4-2.8)^2+(4-6.4)^2} = \sqrt{7.2} \approx \boxed{2.68}$ .

Any point on the curve $y^2=4x$ is of the form $(t^2,2t)$ , where $t \in \mathbb{R}$ is a parameter.

Now the distance of a point $(\alpha,\beta)$ from a straight line $ax+by+c=0$ is given by $\dfrac{|a\alpha+b\beta+c|}{\sqrt{a^2+b^2}}$ .

Therefore, the distance of $(t^2,2t)$ from the line $x-2y+10=0$ will be $\dfrac{t^2-4t+10}{\sqrt{5}}=\dfrac{(t-2)^2+6}{\sqrt{5}}$ , whose minimum occurs at $t=2$ , hence the minimum distance of the curve from the straight line is $\boxed{\dfrac{6}{\sqrt{5}}\approx 2.683}$ .

The distance between a point $(x0,y0)$ and a line $ax+by+c=0$ is $|ax0+by0+c|/√(a^2 + b^2)$ . Since the point is on $x=y^2/4$ , $x0=y0^2/4$ . Thus, $|(y0^2)/4 - 2(y0) + 10|/√(5)$ would be the distance. We can split this into two cases where $d = (y0^2)/4 - 2(y) + 10/√(5)$ and $d=-((y0^2)/4 - 2(y) + 10)/√(5)$ . Since both of these are quadratic functions, the minimum can be found by determining the vertex. The minimum comes out to be the positive answer from the first case, which is $2.683$ .