Find a point on parabola $y^2 =4x$ which is at shortest distance from $(1,0)$.
Answer given is $(0,0)$
But I am getting imaginary values.
How do i get $(0,0)$
Point on parabola which is at shortest distance
calculus
Related Solutions
I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function?
As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function
$$\begin{equation} d(x,y,z)=\sqrt{x^{2}+y^{2}+(z-1)^{2}} \tag{1a} \end{equation}$$
subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$ $$\begin{equation} g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0. \tag{2} \end{equation}$$ Since $\sqrt{x^{2}+y^{2}+(z-1)^{2}}$ increases with $x^{2}+y^{2}+(z-1)^{2}$ you can simplify the computations if you find the minimum of $$\begin{equation} [d(x,y,z)]^2=x^{2}+y^{2}+(z-1)^{2} \tag{1b} \end{equation}$$ subject to the same constraint $(2)$. The Lagrangian function is then defined by $$\begin{eqnarray} L\left( x,y,z,\lambda \right) &=&[d(x,y,z)]^2+\lambda g(x,y,z) \\ L\left( x,y,z,\lambda \right) &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z- \frac{3}{2}\left( x^{2}+y^{2}\right) \right), \tag{3} \end{eqnarray}$$ where $\lambda $ is the Lagrange multiplier. By this method you need to solve the following system $$\begin{equation} \left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0, \frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda } =0,\right. \tag{4} \end{equation}$$ which results in
$$\begin{eqnarray} \left\{ \begin{array}{c} 2x+3\lambda x=0 \\ 2y+3\lambda y=0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} x=0\vee 2+3\lambda =0 \\ y=0\vee 2+3\lambda =0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} x=0 \\ y=0 \\ \lambda =2 \\ z=0 \end{array} \right. \vee \left\{ \begin{array}{c} \lambda =-2/3 \\ z=2/3 \\ x^{2}+y^{2}=4/9 \end{array} \right. \tag{5} \end{eqnarray}$$
For $x=y=x=0$ we get $d(0,0,0)=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions $$\begin{equation} \underset{g(x,y,z)=0}\min d(x,y,z)=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}. \tag{6} \end{equation}$$
It is attained on the intersection of the surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$.
$$\text{Plane }z=\frac{2}{3} \text{(blue) and surface }z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $$
Notes.
- As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find $$\begin{equation} \min [d(r)]^2=r^{2}+(\frac{3}{2}r^{2}-1)^{2} \tag{7} \end{equation}$$ and then find $d(r)=\sqrt{[d(r)]^2}$ at the minimum.
- The surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.
Distance from origin is $\sqrt{x^2+y^2} = \sqrt{x^2+3 x}$. Let $x_0=1$, $h=0.1$; then the change in distance from the origin is
$$\sqrt{(x_0+h)^2+3 (x_0+h)} - \sqrt{x_0^2+3 x_0} \approx h \left [\frac{d}{dx} \sqrt{x^2+3 x} \right ]_{x=x_0} = \frac{2 x_0+3}{2 \sqrt{x_0^2+3 x}}h = 0.125$$
Typically, problems asking for small changes in a function in response to small changes in their arguments are asking for the evaluation of a derivative of the function.
Best Answer
Note : $y^2=4x$ is a parabola, vertex $(0,0)$, symmetric about the $x-$axis.
We have $x \ge 0.$
Distance$^2$:
$d^2= y^2 + (x-1)^2 =4x +(x-1)^2$
$d^2=x^2+2x+1= (x+1)^2.$
$ \min (d^2) =$
$ (0+1)^2 =1$ at $x=0$ (why?)
(Recall $x \ge 0$)
The point on the parabola that minimizes $d^2$ to $(1,0)$ is (0,0)
Note : $d^2$ has been minimized, does this imply $d$ is minimal?