Closest Point On Parabola $x = 2y^2$ To (10, 0)

Hey guys! Today, let's tackle a classic calculus problem: finding the point on a parabola that's closest to a given point. Specifically, we're looking for the point on the parabola $x = 2y^2$ that's nearest to the point $(10, 0)$. This is a fun exercise in optimization, combining geometry and calculus to get to the solution. So, buckle up, and let's dive in!

Setting up the Distance Function

First, let's think about the problem. We need to find a point on the parabola. Any point on the parabola $x = 2y^2$ can be represented as $(2y^2, y)$. We want to minimize the distance between this point and $(10, 0)$. Remember the distance formula? It's our best friend here. The distance, $D$, between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\qquad D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

In our case, $(x_1, y_1) = (2y^2, y)$ and $(x_2, y_2) = (10, 0)$. Plugging these values into the distance formula, we get:

$\qquad D = \sqrt{(10 - 2y^2)^2 + (0 - y)^2}$

$\qquad D = \sqrt{(10 - 2y^2)^2 + y^2}$

Now, to make things a bit easier, we can minimize the square of the distance instead of the distance itself. Why? Because the minimum of $D^2$ occurs at the same $y$-value as the minimum of $D$. This avoids dealing with the square root, which simplifies our calculations. Let's call the square of the distance $f(y)$:

$\qquad f(y) = D^2 = (10 - 2y^2)^2 + y^2$

This $f(y)$ is the function we want to minimize. It represents the squared distance between a point on the parabola and the point (10, 0), expressed in terms of the y-coordinate of the point on the parabola. Minimizing this function will give us the y-coordinate of the point on the parabola that's closest to (10, 0). Remember, squaring the distance doesn't change where the minimum occurs, but it makes the math much simpler. So, by working with $f(y)$ instead of $D$, we avoid dealing with the square root in our calculations. This is a common trick in optimization problems, and it can save you a lot of time and effort.

Minimizing the Distance Function

Okay, guys, we've got our distance function squared: $f(y) = (10 - 2y^2)^2 + y^2$. Now, it's time to put on our calculus hats and find the minimum of this function. How do we do that? You guessed it – derivatives! We'll find the critical points of $f(y)$ by taking its derivative, setting it equal to zero, and solving for $y$.

First, let's expand $f(y)$ to make it easier to differentiate:

$\qquad f(y) = (100 - 40y^2 + 4y^4) + y^2$

$\qquad f(y) = 4y^4 - 39y^2 + 100$

Now, let's find the first derivative, $f'(y)$:

$\qquad f'(y) = 16y^3 - 78y$

To find the critical points, we set $f'(y) = 0$:

$\qquad 16y^3 - 78y = 0$

We can factor out a $2y$:

$\qquad 2y(8y^2 - 39) = 0$

This gives us three possible solutions for $y$:

• $y = 0$
• $8y^2 - 39 = 0 \Rightarrow y^2 = \frac{39}{8} \Rightarrow y = \pm \sqrt{\frac{39}{8}} = \pm \frac{\sqrt{78}}{4}$

So, our critical points are $y = 0$, $y = \frac{\sqrt{78}}{4}$, and $y = -\frac{\sqrt{78}}{4}$. These are the potential $y$-values where our function $f(y)$ might have a minimum (or a maximum). To confirm that these are indeed minima, we can use the second derivative test. Let's find the second derivative, $f''(y)$:

$\qquad f''(y) = 48y^2 - 78$

Now, we'll evaluate $f''(y)$ at each critical point:

• For $y = 0$: $f''(0) = -78 < 0$. This indicates a local maximum.
• For $y = \frac{\sqrt{78}}{4}$: $f''(\frac{\sqrt{78}}{4}) = 48(\frac{39}{8}) - 78 = 234 - 78 = 156 > 0$. This indicates a local minimum.
• For $y = -\frac{\sqrt{78}}{4}$: $f''(-\frac{\sqrt{78}}{4}) = 48(\frac{39}{8}) - 78 = 234 - 78 = 156 > 0$. This also indicates a local minimum.

So, we have local minima at $y = \frac{\sqrt{78}}{4}$ and $y = -\frac{\sqrt{78}}{4}$. These are the y-coordinates of the points on the parabola closest to (10, 0).

Finding the Points on the Parabola

Alright, awesome work, guys! We've found the $y$-coordinates of the points on the parabola closest to $(10, 0)$. Now, we need to find the corresponding $x$-coordinates. Remember the equation of our parabola: $x = 2y^2$. We just need to plug in our $y$-values to find the $x$-values.

Let's start with $y = \frac{\sqrt{78}}{4}$:

$\qquad x = 2(\frac{\sqrt{78}}{4})^2 = 2(\frac{78}{16}) = \frac{78}{8} = \frac{39}{4}$

So, one point is $(\frac{39}{4}, \frac{\sqrt{78}}{4})$.

Now, let's do the same for $y = -\frac{\sqrt{78}}{4}$:

$\qquad x = 2(-\frac{\sqrt{78}}{4})^2 = 2(\frac{78}{16}) = \frac{78}{8} = \frac{39}{4}$

We get the same $x$-coordinate! So, our other point is $(\frac{39}{4}, -\frac{\sqrt{78}}{4})$.

Therefore, the points on the parabola $x = 2y^2$ that are closest to the point $(10, 0)$ are:

• $(\frac{39}{4}, \frac{\sqrt{78}}{4})$
• $(\frac{39}{4}, -\frac{\sqrt{78}}{4})$

These are the points where the distance to $(10, 0)$ is minimized. You can even visualize this – imagine the parabola opening to the right, and the point $(10, 0)$ sitting to the right of the vertex. It makes sense that there would be two points on the parabola, symmetrically placed above and below the x-axis, that are closest to $(10, 0)$.

Wrapping it Up

So, there you have it! We successfully found the points on the parabola $x = 2y^2$ that are closest to the point $(10, 0)$. We used the distance formula, minimized the square of the distance to simplify calculations, found critical points using derivatives, and confirmed our minima using the second derivative test. This problem beautifully illustrates the power of calculus in solving real-world optimization problems. Remember guys, practice makes perfect, so keep those calculus skills sharp! You've got this! This was a great example of using calculus to solve a geometric problem. By translating the geometric idea of distance into a function that we could minimize, we were able to find the exact points on the parabola that are closest to the given point. This approach is applicable to many different types of optimization problems, so it's a valuable tool to have in your mathematical toolkit. Keep exploring, keep learning, and keep having fun with math!