Question

Dada la ecuacion 25x2 + 4y2 = 100, determina las coordenadas de los vertices, focos, las longitudes de los respectivos ejes mayor y menor, la excentricidad, la longitud de los lados rectos y realiza la representacion grafica
URGENTE

Answer 1

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.