Understanding the Length of Axes in an Ellipse Equation

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In the discussion about the ellipse equation, the standard form is identified as (x^2/a^2) + (y^2/b^2) = 1, where 'a' and 'b' represent the lengths of the semi-major and semi-minor axes, respectively. The coefficients dictate the x-intercepts and y-intercepts, which correspond to the lengths of these axes. When x=0, the equation reveals the y-values as ±b, and when y=0, the x-values are ±a, establishing the ellipse's dimensions. The relationship between 'a' and 'b' determines the shape of the ellipse, with equal values resulting in a circle. Understanding these relationships clarifies how the lengths of the axes are derived from the equation.
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Homework Statement


As I was relearning some concepts in calculus, I came across a section on ellipses. What I don't understand is why a and b in the standard equation of an ellipse govern the length of the minor/major axes. Can anyone shed some light? Thank you very much!


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The Attempt at a Solution


I have not attempted a solution, only tried to visualize the results in my head.
 
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Can you show us which equation you mean when you say 'Standard Equation', as there are several different ways which the equation of the ellipse can be presented. What you refer to as 'a and b' needs some context.

Either way, the co-efficient of x (when y = 0) dictates the x-intercepts (this distance between the two x points represents one axis) and the co-efficient of y (when x = 0) dictates the y-intercepts (this distance between the two y points represents the other axis). You get two points in each instance of course because to solve for x or y you have to take a square root on both sides of the equation, and thus get a +/- number.

A fairly common ellipse is 9x^2 + 4y^2 = 36. Solve for x (when y = 0) and then solve for y (when x = 0) and then plot the 4 points and join them to see the ellipse and its major and minor axes.

The special case of these distances being equal occurs when the ellipse is a circle.

Other equations involve x^2 and y^2 being fractions and always equaling 1. For example x^2/4 + y^2/9 = 1 is the same as 9x^2 + 4y^2 = 36 if you rearrange things.
 
Thank you for the quick reply! The equation I was referring to was (x^2/a^2)+(y^2/b^2)=1.
 
That is the equation of an ellipse with center at (0, 0) and with its axes of symmetry along the x and y axes.

Look at what happens if x or y is 0.

If x= 0, then 0^2/a^2+ y^2/b^2= y^2/b^2= 1 so y^2= b^2 and y= \pm b.

If y= 0, then x^2/a^2+ 0^2/b^2= x^2/a^2= 1 so x^2= a^2 and x= \pm a.

On the other hand, if x is not 0, then, since a square is never negative, y^2/b^2 must be less than 1 so y must be between -b and b. If y is not 0 then x^2/a^2 must be less than 1 so x must be between -a and a. That is, the ellipse goes form (-a, 0) to (a, 0) on the x-axis and from (0, -b) to (b, 0) on the y-axis.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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