How do you find the standard form of $3 \frac{x^{2}}{12} + 5 \frac{y^{2}}{500} = 1$ $3x^2/12+ 5y^2/500 = 1$ and what kind of a conic is it?

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Answer 1 · 2016-01-04T23:38:57

An ellipse.

Simplify the fractions.

$\frac{x^{2}}{4} + \frac{y^{2}}{100} = 1$ $x^2/4+y^2/100=1$

This fits the mold for an ellipse, which has the general form

$\frac{{(y - k)}^{2}}{a^{2}} + \frac{{(x - h)}^{2}}{a^{2}} = 1$ $(y-k)^2/a^2+(x-h)^2/a^2=1$

However, the terms at first are switched since $a > b$ $a>b$ for all ellipses, here denoting a vertically oriented ellipse.

Thus, the actual general form is

$\frac{y^{2}}{100} + \frac{x^{2}}{4} = 1$ $y^2/100+x^2/4=1$

Here, $h = 0, k = 0, a = 10, b = 2$ $h=0,k=0,a=10,b=2$ .

graph{y^2/100+x^2/4=1 [-22.81, 22.8, -11.4, 11.41]}

How do you find the standard form of 3x212+5y2500=13x^2/12+ 5y^2/500 = 1 and what kind of a conic is it?