Question

III CONIC SECTIONS Finding the center, vertices, and foci of a v Find the center, vertices, and foci of the ellipse. Simplify your answers as much as possible. (x+4)2 (y-2)2 + =1 100 64 Center: 00 Vertices: (01) and (0) Foci: (1) and CD

Answer 1

To answer this question, first we must need to know the general equation of an ellipse. It is given by:

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

In the given equation the ellipse is centered about the point (h,k). Any standard ellipse i.e. with the x-y axes as it's major and minor axes can be modelled in this form. So, if we look at the given question, it is similar to the equation given here. Hence, if we determine the values of h,k,a, and b, then all information about the ellipse can be obtained.

As mentioned earlier (h,k)represent the center of the ellipse. There are two axes in an ellipse: the major and the minor axes. In the equation considered here the x and y axes form the axes and if b<a, then x axis is the major axis while y axis is the minor one and vice versa. As shown in the following figure we have to determine the center, vertices and foci.

у Co-vertex Minor axis Vertex Focus Focus Vertex X Major axis Center Co-vertex

It can be done if we see the following diagram:

A(0, b) Minor axis (-a, 0) (c, 0) (-0,0) Major axis (0,0) (a,0) -X *(0-b)

Thus, we can determine all the required terms once the parameters are obtained. For that we compare the given ellipse equation with the known standard one. In both of the equations, there are two square terms adding up on the left side resulting in 1 in the right side. Furthermore, from the two square terms one is in x and the other in y in both the equations. To be able to obtain the values we see the two equations again:

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

$\frac{(x+4)^{2}}{100} + \frac{(y-2)^{2}}{64} = 1$

This equation can be modified as follows:

$\frac{(x-(-4))^{2}}{10^{2}} + \frac{(y-(2))^{2}}{8^{2}} = 1$

From here we can say that: h = -4, k = 2, a = 10,and b = 8.

Using these values the center is known as (h,k) = (-4,2). In the figure shown the vertices are at (a,0) and (-a,0), but it is for an ellipse with center (0,0). If the center is (h,k) then the vertices shift to (h+a,k) and (h-a,k). So the vertices of this ellipse are given by: (-14,0) and (6,0).

For the foci, we need to determine c. The relation between c, a, b is given by:

So, $c = \sqrt{100-64} = 6$

And similar to the vertices the foci must also shift by (h,k).

Hence the foci are: (h+c,k) and (h-c,k)  i.e. (2,2) and (-10,2).