1/4x^3 = -27/4 can someone help me

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photon22

05.09.2020

Mathematics

The standard form of the ellipse is $\frac{x^{2}}{25} +\frac{y^{2}}{169} = 1$ . Coordinates of the foci are $F_{1} (x,y) = (0,-c)$ and $F_{2}(x,y) = (0,c)$ , respectively.

Step-by-step explanation:

The equation of a ellipse centered in the origin in standard form is defined by following formula:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$ (1)

Where a, b are the coefficients of the ellipse.

Now we proceed to transform the equation of the ellipse in general form into standard form by algebraic means:

$169\cdot x^{2}+25\cdot y^{2}-4225 = 0$

$169\cdot x^{2}+25\cdot y^{2} = 4225$

$\frac{x^{2}}{25} +\frac{y^{2}}{169} = 1$

Since b a , then major axis of the ellipse is located in y-axis. The distance between center and focus () is calculated by following Pythagorean identity:

$c =\sqrt{b^{2}-a^{2}}$ (2)

c = 12

The location of the foci are represented by $F_{1} (x,y) = (0,-c)$ and $F_{2}(x,y) = (0,c)$ . If we know that c = 12 , then the location of the foci are, respectively:

$F_{1}(x,y) = (0,-12)$ , $F_{2}(x,y) = (0,12)$

Put the equation 169x^2+25y^2-4225=0 in standard form and sketch the ellipse including the foci in y

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