Animated Solution for Mathematics - Conic Sections: If the point on the curve $y^2=6x$, nearest to the point $(3,\frac{3}{2})$ is $(\alpha ,\beta )$, then $2(\alpha +\beta )$ is equal to

The shortest distance from a point $Q$ to a curve is always found along the normal line at the point $P$ on the curve. If we imagine a circle centered at $Q$ expanding until it touches the parabola, the point of contact $P$ occurs where the radius of the circle (the normal to the parabola) meets the curve.

Our target point is $Q(3,\frac{3}{2})$ and the parabola is defined by $y^2=6x$. Our objective is to find the coordinates $P(\alpha ,\beta )$ such that the normal at $P$ passes through $Q$.

The standard form of a parabola is $y^2=4ax$. Comparing this to $y^2=6x$, we identify $4a=6$, which yields:

To simplify the calculations, we represent any point $P$ on the parabola using the parameter $t$ as $P(a{t}^2,2at)$. Substituting our value of $a$, the coordinates of $P$ become:

The standard parametric equation for the normal to a parabola at point $t$ is given by:

Substituting $a=\frac{3}{2}$ into this equation, we obtain the general form for the normal line:

Since the normal must pass through $Q(3,\frac{3}{2})$, we substitute $x=3$ and $y=\frac{3}{2}$ into the equation:

Observe that the term $3t$ appears on both sides of the equation. Subtracting $3t$ from both sides results in a significant simplification:

Dividing both sides by $\frac{3}{2}$, we find $t^3=1$. The only real solution for this equation is:

Now that we have determined $t=1$, we substitute this value back into our parametric coordinates for $P$:

Thus, the point $P$ is $(\frac{3}{2},3)$. The problem asks for the value of $2(\alpha +\beta )$:

The final result is 9.