Question

Write the equation of the ellipse with center at (21), one vertex at (2-4), and one focus at (2,-2)

Answer 1

Answer:

$\frac{(x-2)^2}{16} +\frac{(y-1)^2}{25} =1$

Step-by-step explanation:

The equation of an ellipse outside the origin is as follows

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

The above is for an ellipse centered on (h, k). in this case the center is in (2,1) so $h = 2$ and $k = 1$

So far we have:

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

Now we find a and b, Where b is the semi-major axis and the minor semi-axis.

If the center is at (2, 1) and one focus is at (2,-2) this means there are 3 units of distance between the center and the focus. This quantity will be called c.

$c=3$

Now, of one vertex is at (2,-4) this means there are 5 units of distance between the center and the vertex, this is the semi major axis of the ellipse

$b= 5$ ⇒ $b^2=5^2=25$

and to find a:

$b^2=a^2+c^2$

clearing for a:

$a^2=b^2-c^2$

$a^2=5^2-3^2$

$a^2=25-9$

$a^2=16$

Substituting in the equation of the ellipse

$\frac{(x-2)^2}{16} +\frac{(y-1)^2}{25} =1$

wich is the second of the options.

Answer 2

B. (x-2)^2/16 + (y-1)^2/25=1