# Finding closest point from a parabola to a plane

1. Dec 6, 2013

### mahler1

The problem statement, all variables and given/known data.

Find the point in the parabola $y^2=x$, $z=0$ closest to the plane $z=x+2y+8$

The attempt at a solution.

I've solved some problems where I had to find the closest point from a given surface to another point on the space. In this case, I have to find the distance between a curve and a plane so to speak. Maybe I could use Lagrange multipliers, but I am not so sure how to apply it in this particular problem.

2. Dec 6, 2013

### ShayanJ

You can use Lagrange multipliers but the tricky part is finding the function which should be minimized.That can be done by recalling the definition of the distance of a point to a plane. That's just the length of the line from the point normal to the plane.The normal vector of your plane is $\hat{n}=\frac{1}{\sqrt{5}}(1,2,-1)$. Now consider a point $P(x,y,z)$ in space.But I want P to be on the parabola you mentioned so it becomes $P(y^2,y,0)$.I make a line with P and $-\hat{n}$.$L:(y^2-tn_x,y-tn_y,-tn_z)$ t being a parameter.Now lets see in what t,my line intersects your pane!
$-tn_z=y^2-tn_x+2(y-tn_y)+8 \Rightarrow (n_x+2n_y-n_z)t=y^2+2y+8 \Rightarrow t=\frac{y(y+2)+8}{n_x+2n_y-n_z}=t_p$
Now we can find the distance of a point on your parabola to your plane by $d=\sqrt{(y^2-t_p n_x-y^2)^2+(y-t_p n_y-y)^2+(-t_pn_z)^2}=t_p \sqrt{n_x^2+n_y^2+n_z^2}$ But $\hat{n}$ is a unit vector so $d=t_p$.The function you should minimize,is $t_p$.
But wait...We now have a function of one variable with no constraints...so you can just differentiate with respect to $y$ and set it equal to zero and that gives you the answer.

Last edited: Dec 6, 2013
3. Dec 6, 2013

### HallsofIvy

Staff Emeritus
Another way to do it is to use the geometric fact that the "shortest distance" to a plane is always along a line perpendicular to the plane. We can write this plane as x+ 2y- z= -8 so it has <1, 2, -1> as perpendicular vector. A line through $(x_0, y_0, 0)$ in the direction of that vector is $x= x_0+ t$, $y= y_0+ 2t$, $z= -t$. Further we have $y_0= x_0^2$ so we can write that line as $x= x_0+ t$, $y= x_0^2+ 2t$, $z= -t$.

That line will cross the plane when $x_0+ 2(x_0^2+ 2t)- (-t)= 5t+ x_0+ 2x_0^2= -8$. You can solve that for t in terms of $x_0$, so determine the point in terms of $x_0$. Then minimize the distance (more easily, the distance squared) to find $x_0$.

4. Dec 6, 2013

### Ray Vickson

Besides what others have said, you can (as you wanted) use the Lagrange multiplier method: minimize the squared distance from (t^2,t,0) to (x,y,z), subject to x+2y-z+8=0. The variables are t,x,y,z. This works out very nicely.

Last edited: Dec 6, 2013