# Thread: Symmetry of y = sqrt(2 - a^2)

1. ## Symmetry of y = sqrt(2 - a^2)

Hi!

i am trying to determine if this equation is symmetrical to the x-axis and/or the y-axis. or both. the equation is...

$y=\sqrt{2-a^2}$

which i changed to...

$y^2=2-a^2$

the book states that this equation is ONLY symmetrical to the y-axis. i agree it is symmetrical to the y-axis, but it also seems to me to be symmetrical to the x-axis.

since...

$(-y)^2=2-a^2 \Rightarrow y^2=2-a^2$

Because a square root has both a positive and negative value, doesn't this make sense?

what am i missing/misunderstanding?

p.s. is this algebra or pre-calc? i wasn't entirely sure

2. Note that your expression is only real for $-\sqrt{2} \leq a \leq \sqrt{2}$

Underneath the radical it's the difference of two squares, which iirc, are always symmetrical in the y axis. Plus your graph is a semi-circle.

No doubt someone can put it more elegantly than I though

3. Originally Posted by e^(i*pi)
Note that your expression is only real for $-\sqrt{2} \leq a \leq \sqrt{2}$

Underneath the radical it's the difference of two squares, which iirc, are always symmetrical in the y axis. Plus your graph is a semi-circle.

No doubt someone can put it more elegantly than I though
thanks. but i still don't get it.

is $y=\sqrt{2-a^2}$ another expression of $y^2=2-a^2$

edit: ok obviously it's not.

something about the a^2 under the radical i gather. suppose i wanted to square $y=\sqrt{2-a^2}$ ...how would i do this?

4. ok lets say a=.5

then...

$y=\sqrt{2-.25}\approx1.3229 and 1.3229^2=1.75=y^2=2-.25$

is this totally wrong?

my graphing calculator does show that it is a semi-circle.

edit: ok, so you can never have a negative number as the answer for a square root. so is this just something that one has to remember? i guess i'm struggling with the fact that the method i learned from the book i'm reading explained how to determine symmetry using specific rules.

it states that you can substitute in -y or -b in place of y or b respectively when trying to see if a graph or function displays symmetry with the x-axis. (y=b). If the resulting equation is the same as the original, then it is symmetrical with respect to the x-axis.

so when i plug in -y i get $-y=\sqrt{2-a^2}$

so is it as simple as saying that $-y\neq y$ ?

this inequality isn't always true though, since y could be zero.

5. Originally Posted by jonnygill
thanks. but i still don't get it.

is $y=\sqrt{2-a^2}$ another expression of $y^2=2-a^2$

edit: ok obviously it's not.

something about the a^2 under the radical i gather. suppose i wanted to square $y=\sqrt{2-a^2}$ ...how would i do this?
You correctly squared the equation, the issue with squaring is that it introduces possible extraneous solutions (solutions not in the original domain). For example if you had $y = 2$ and squared it you'd get $y^2 =4$. Solutions here are $\pm 2$

In your case if you wanted to solve $y^2 = 2-a^2$ for y then you'd have $y = \pm \sqrt{2-a^2}$ and the negative solution is extraneous.

Your equation will hold for $|a| \leq \sqrt{2}$

Using your example of a = 0.5

$y = \sqrt{2-0.5^2} = \dfrac{\sqrt{7}}{2}$ and $y^2 = 2-0.5 = \dfrac{7}{4} = \left(\dfrac{\sqrt{7}}{2}\right)^2$

Say if you were only given the second equation $y^2 = \dfrac{7}{4}$ then you'd deduce that $y = \pm \dfrac{\sqrt{7}}{2}$ - but where has that negative solution come from?

Wiki: http://en.wikipedia.org/wiki/Extrane...sing_solutions

6. i guess it would come from squaring the radical and then taking the square root of the answer you get.

7. Originally Posted by jonnygill
Hi!

i am trying to determine if this equation is symmetrical to the x-axis and/or the y-axis. or both. the equation is...

$y=\sqrt{2-a^2}$

which i changed to...

$y^2=2-a^2$

the book states that this equation is ONLY symmetrical to the y-axis. i agree it is symmetrical to the y-axis, but it also seems to me to be symmetrical to the x-axis.

since...

$(-y)^2=2-a^2 \Rightarrow y^2=2-a^2$

Because a square root has both a positive and negative value, doesn't this make sense?

what am i missing/misunderstanding?

p.s. is this algebra or pre-calc? i wasn't entirely sure
Your $y$ is a constant depending on the parameter $a$, so assuming $a$ is such that $y$ is real the curve is a horizontal line in the $x-y$ plane, and so symmetric about the $y$ axis

CB

8. Squaring both sides of an equation in general is a "non-reversible" step. When you "undo" the squaring operation you don't get back what you started with. For example, if $x=3$, and you square both sides you get $x^2=9$. Reversing this gives $x=\pm 3$. An extra solution has been generated that wasn't there before.