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I am given the equation: $9x^2 + 4y^2 + z = 3$

The standard equation of a paraboloid parallel to z-axis is: $$\frac{z-z_0}{c} = \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2}$$ I think the given equation should be a paraboloid due to the $z$ component, then I try to put it in a std. form.

$9x^2+ 4y^2 + z = 3$

$\frac{9x^2}{3} + \frac{4y^2}{3} + \frac{z}{3} = 1$

$3x^2 + \frac{4}{3}y^2 = 1 - \frac{z}{3}$

And then, I am confused since this is obviously not the std form. The one and also the negative value of z is not std.

I would appreciate any tips or someone pointing me the right way!

Certainly not a dog
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Amai
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1 Answers1

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Note $1 - \frac{z}{3} = \frac{3 - z}{3} = \frac{z - 3}{-3}$. Thus, $z_0 = 3$ and $c = -3$, along with $x_0 = y_0 = 0$, $a = \frac{1}{\sqrt{3}}$ and $b = \frac{\sqrt{3}}{2}$, allows your equation to be in standard form.

John Omielan
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  • I feel like as I progress in math, all the prior things are slowly fading away...just to confirm, doing something to the z side as you have shown me, does not affect the x-y side right? – Amai Sep 16 '19 at 05:19
  • @Amai All I'm doing on the $z$ side is expressing it differently. Since I'm not adding, subtracting, dividing or multiplying the result by any value, it doesn't change it's overall value. Thus, what I've one doesn't affect the $x-y$ side value, so there's nothing you need to do to it to ensure the equality still holds in the equation. – John Omielan Sep 16 '19 at 05:21
  • Understood! One last thing that I'm a bit confused about. Can c be a negative value? – Amai Sep 16 '19 at 05:23
  • @Amai The $c$ is only to the first power, so there's no square root involved. Also, there's not requirement for $c$ to be positive. Note if $c$ is negative compared to using a positive value of $-c$ means, for any given $x,y$, the $z - z_0$ either stays at $0$ or switches sign between positive and negative. The resulting paraboloid is a refection in the plane $z = z_0$ of the paraboloid you would get if $c \gt 0$. – John Omielan Sep 16 '19 at 05:24
  • Yes I realized that so I editted by question (but too late i guess). Um...is it possible to explain how a negative c value differs from a pos. one. I think it is just whether it opens up or down, is that correct? – Amai Sep 16 '19 at 05:26
  • @Amai Yes, as I commented above about positive vs. negative values being a reflection in the plane $z = z_0$, the sign of $c$ does control whether it opens up or down, as you ask. – John Omielan Sep 16 '19 at 05:28
  • Thanks a bunches! I really appreciate the help. It seems that the simple things always gets me nowadays. – Amai Sep 16 '19 at 05:31