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Alex_Xolod [135]
3 years ago
14

Dada la ecuacion 25x2 + 4y2 = 100, determina las coordenadas de los vertices, focos, las longitudes de los respectivos ejes mayo

r y menor, la excentricidad, la longitud de los lados rectos y realiza la representacion grafica
URGENTE

Mathematics
1 answer:
Likurg_2 [28]3 years ago
7 0

Answer:

The given equation is

25x^{2} +4y^{2}=100

Which represents an elipse.

To find its elements, we need to divide the equation by 100

\frac{25x^{2} +4y^{2} }{100} =\frac{100}{100} \\\frac{x^{2} }{4} +\frac{y^{2} }{25} =1

Where a^{2} =25 and b^{2}=4. Remember that the greatest denominator is a, and the least is b. So, we extract the square root on each equation.

a=5 and b=2.

In a elipse, we have a major axis and a minor axis. In this case, the major axis is vertical and the minor axis is horizontal, that means this is a vertical elipse.

The length of the major axis is 2a=2(5)=10.

The length of the minor axis is 2b=2(2)=4.

The vertices are (0,5);(0,-5) and (2,0);(-2,0).

Now, the main parameters of an elipse are related by

a^{2}=b^{2} +c^{2}, which we are gonna use to find c, the parameter of the focus.

c=\sqrt{a^{2}-b^{2} }=\sqrt{25-4}=\sqrt{21}

So, the coordinates of each focus are (0,\sqrt{21}) and (0,-\sqrt{21})

The eccentricity of a elipse is defined

e=\frac{c}{a}=\frac{\sqrt{21} }{5}  \approx 0.92

The latus rectum is defined

L=\frac{2b^{2} }{a}=\frac{2(4)}{5} =\frac{8}{5} \approx 1.6

Finally, the graph of the elipse is attached.

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