Animated Solution for Mathematics - Conic Sections: The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is :

To find the shortest distance between the parabola $y^2=8x$ and the circle $x^2+y^2+12y+35=0$, we must first identify the geometric properties of both curves.

The parabola $y^2=8x$ is in the standard form $y^2=4ax$, where $4a=8$, implying $a=2$.

For the circle, we complete the square for the equation $x^2+y^2+12y+35=0$:

Thus, the circle is centered at $C(0,-6)$ with a radius $r=1$.

The shortest distance between two non-intersecting curves lies along their common normal. For a circle, any normal line must pass through its center $C(0,-6)$.

Therefore, we must find a normal to the parabola $y^2=8x$ that passes through the point $C(0,-6)$.

The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by:

Substituting $a=2$ into this equation, we obtain:

Since this normal passes through $C(0,-6)$, we substitute $x=0$ and $y=-6$:

By inspection, we find that $m=1$ is a root of this cubic equation.

With the slope $m=1$, we determine the point of contact $P$ on the parabola using the coordinates $(a{m}^2,-2am)$:

Next, we calculate the distance $PC$ between the center $C(0,-6)$ and the point $P(2,-4)$:

The shortest distance $d$ between the parabola and the circle is the distance from the center to the parabola minus the radius of the circle:

The minimum distance between the two curves is $2\sqrt{2}-1$.