Shortest distance between the origin and a parabola

Question

What is the smallest distance between the origin and a point on the graph of $y = \frac{1}{\sqrt{2}} (x^2 - 18)?$

I used the distance formula to get $$\sqrt{x^2 + \frac{1}{2}(x^2-18)^2},$$ but I don't know what to do from there.

Answer 1

There are 2 ways you can solve it. One using basic algebra and other requires calculus.

I) Completing the Square (Algebra)-

This is a classic use case of the completing the square method of factorisation, which is also seen quite a bit in inequalities. So I suggest you try to understand it well.

Here’s how you can approach this question- $$ Distance = \sqrt{x^2+\frac{1}{2}(x^2-18)^2} = \sqrt{x^2+\frac{1}{2}(x^4-36x^2 +18^2)} \\ = \sqrt{\frac{1}{2}(x^4)-17x^2 +\frac{1}{2}18^2} $$ Now we complete the square, by adding or subtracting appropriate terms, such that, $a^2 = \frac{1}{2}x^4, -2ab = -17x^2$ so that we can use the formula $(a-b)^2 = a^2 - 2ab + b^2$ $$ \implies \sqrt{{(\frac{1}{\sqrt{2}}}x^2)^2 - 2(\frac{1}{\sqrt{2}}x^2)(\frac{17}{\sqrt{2}}) + (\frac{17}{\sqrt{2}})^2 - (\frac{17}{\sqrt{2}})^2 +\frac{1}{2}18^2} \\ \implies \sqrt{{(\frac{1}{\sqrt{2}}}x^2 - \frac{17}{\sqrt{2}})^2 + \frac{18^2}{2} - \frac{17^2}{2}} = \sqrt{{(\frac{1}{\sqrt{2}}}x^2 - \frac{17}{\sqrt{2}})^2 + \frac{35}{2}} $$

Now the answer lies in the observation that ${(\frac{1}{\sqrt{2}}}x^2 - \frac{17}{\sqrt{2}})^2 \geq 0$, This is because, the square of any real number is never negative, and the least value a perfect square can attain is $0$. This is a very useful tool in many questions, as it allows us to introduce inequalities in places where inequalities haven't been given (So questions like find the maximum or minimum) Now we may substitute ${(\frac{1}{\sqrt{2}}}x^2 - \frac{17}{\sqrt{2}})^2 = 0$ into our equation to get the least value of distance. $$ {Distance}_{min} = \sqrt{\frac{35}{2}} $$ Which is the answer and can be verified.

I will not detail the calculus approach here, as I am not sure if you are familiar with calculus. However, if you want me to, you may ask me in the comments. Hope it helps!

Answer 2

An alternative approach is as follows:

The shortest distance is the radius of the smallest circle $$x^2+y^2=r^2$$ that has point of intersection with the parabola $$y=\frac{1}{\sqrt 2}(x^2-18).$$

Putting the equations together, we have

$$x^2+\frac{1}{2}(x^2-18)^2=r^2$$ $$x^4-34x^2+(324-2r^2)=0$$

The circle intersects the parabola when $$34^2-4(1)(324-2r^2) \geq 0$$

$$r^2 \geq \frac{35}{2}$$

$$r \geq \sqrt {\frac{35}{2}}$$

Therefore the shortest distance is $\sqrt {\frac{35}{2}}$.

Answer 3

The parametric equation of the parabola is

$ p(t) = ( t, \dfrac{1}{\sqrt{2}} (t^2 - 18) ) $

The squared distance from the origin to $p(t)$ is

$ s = p(t) \cdot p(t) $

Differentiating this with respect to $t$ gives us

$ s' = 2 p(t) \cdot p'(t) $

This must be zero.

$ p'(t) = (1, \sqrt{2} t ) $

Therefore, we want

$ t + t (t^2 - 18) = 0 $

i.e.

$ t (t^2 - 17) = 0 $

The solutions are $ t = 0 $ and $ t = \pm \sqrt{17} $

Now,

$p(0) = (0, - 9 \sqrt{2} ) $ and $ p( \sqrt{17} ) = ( \sqrt{17} , - \dfrac{1}{\sqrt{2}} )$

Therefore,

$s(0) = 162 $ and $ s(\sqrt{17}) = 17.5 $

Hence, the minimum distance is achieved at $t = \pm \sqrt{17} $ and that minimum distance is

$ d = \sqrt{ s (\sqrt{17} ) } = \sqrt{ 17.5 } = \sqrt{\dfrac{35}{2}} $

The two corresponding points on the parabola are

$ ( - \sqrt{17} , - \dfrac{1}{\sqrt{2}} ) $ and $ ( \sqrt{17} , - \dfrac{1}{\sqrt{2}}) $

Answer 4

You can get rid of the square root by observing that out is monotonic over the positive reals. So the square root as a whole is minimal at the point(s) where the expression under the root is minimal.

Now this leaves you with an quartic (degree 4) expression in $x$. But since it only contains even powers of $x$ it's easier to essentially treat $x^2$ as a variable. You can use a new symbol for that, $u:=x^2$. So you have a quadratic expression in $u$ and want to know where that is minimal. That's the case where the derivative is zero. The derivative if a quadratic (degree 2) expression is linear (degree 1) so you get a single value $u$ which satisfies that condition. Assuming it's positive, you then get two values $x_{1,2}=\pm\sqrt u$ which both have the same minimal distance.

Answer 5

Let $P=(x,y)$ is an arbitrary point on the given curve. For the distance between that point and the origin $O=(0,0)$ will hold $$d^2(P,O)=(x-0)^2+(y-0)^2=x^2+y^2 = \sqrt2y+18+y^2,$$

and the minimum is achieved for $y=-\dfrac{1}{\sqrt 2}$, i.e. at the point $P=\left(\sqrt{17},-\dfrac{1}{\sqrt 2}\right)$. Hence $$d=\sqrt{17+\frac{1}{2}}=\boxed{\sqrt{\frac{35}{2}}}.$$

Answer 6

Yet another solution. First calculate all points for which the ray from $O$ is perpendicular to the function, i.e., $$\frac{y}{x}\cdot y'=-1\iff yy'=-x.$$ In our case we have $$\sqrt2\cdot x\frac{1}{\sqrt{2}}(x^2-18)=-x,$$ which immediately gives $x^2=17$ or $x=0$ and $y=-1/\sqrt2$ or $y=-18/\sqrt2$, resp. Hence $$x^2+y^2=17+\frac12\text{ or }17+162 .$$