Question

Help me fast please

Answer 1

Answer:

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

If we compare this to the general expression for an ellipse given by:

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

We can see that the vertex is $V=(0,0)$

And we can find the values of a and b like this:

$a=\sqrt{169}=13, b=\sqrt{25}=5$

in order to find the foci we can find the value of c

$c =\sqrt{169-25}=\sqrt{144}=12$

The two focis are (12,0) and (-12,0)

The convertices  for this case are: (13,0) and (-13,0) on the x axis

And for the y axis (0,5) and (0,-5)

Step-by-step explanation:

For this problem we have the following equation given:

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

If we compare this to the general expression for an ellipse given by:

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

We can see that the vertex is $V=(0,0)$

And we can find the values of a and b like this:

$a=\sqrt{169}=13, b=\sqrt{25}=5$

in order to find the foci we can find the value of c

$c =\sqrt{169-25}=\sqrt{144}=12$

The two focis are (12,0) and (-12,0)

The convertices  for this case are: (13,0) and (-13,0) on the x axis

And for the y axis (0,5) and (0,-5)