Chapter 10, Problem 31RE

To determine

Tofind:the equation of conic using its given description and graph the equation.

Answer to Problem 31RE

The elliptical equation is $\frac{{\left(\text{x+4}\right)}^{\text{2}}}{16}\text{+}\frac{{\left(\text{y-5}\right)}^{\text{2}}}{25}\text{=1}$ .

Explanation of Solution

Given:

Ellipse,

Foci at $\left(-4,2\right)\text{ and }\left(-4,8\right)$ and vertex $\left(-4,10\right)$ .

Concept used:

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

The point $\left(\text{h,k}\right)$ is center of the ellipse.

The coordinates of the vertices are $\left(\text{±a,0}\right)$

The coordinates of the co-vertices are $\left(\text{0,±b}\right)$

The coordinates of the foci are $\left(\text{±c,0}\right)$ where ${\text{c}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{-b}}^{\text{2}}$ .

The length of major and minor axis’s: $\text{2a and 2b}$ respectively.

Where $\text{a>b}$ .

Calculation:

According to the given:

Ellipse,

Foci at $\left(-4,2\right)\text{ and }\left(-4,8\right)$ and vertex $\left(-4,10\right)$

It resembles ellipse in the form:

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

vertex= $\left(\text{h+a, k}\right)\text{=}\left(\text{2,2}\right)$

thus, $\text{a=3}$ .

foci= $\left(\text{h, k}\pm \text{c}\right)$ .

here, $\text{h=-4,k=5 and c=3}$ .

Now vertex is $\left(\text{h,k+a}\right)=\left(-4,10\right)$ .

Thus $\text{a=5}$ .

Now $\text{b=}\sqrt{{\text{a}}^{\text{2}}{\text{-c}}^{\text{2}}}\text{=}\sqrt{\text{25-9}}\text{=4}\text{.}$

The equation of the ellipse is:

  $\frac{{\left(\text{x+4}\right)}^{\text{2}}}{16}\text{+}\frac{{\left(\text{y-5}\right)}^{\text{2}}}{25}\text{=1}$

The equation can be in shown in graph as:

The graph of elliptical equation $\frac{{\left(\text{x+4}\right)}^{\text{2}}}{16}\text{+}\frac{{\left(\text{y-5}\right)}^{\text{2}}}{25}\text{=1}$ .

Hence, elliptical equation is $\frac{{\left(\text{x+4}\right)}^{\text{2}}}{16}\text{+}\frac{{\left(\text{y-5}\right)}^{\text{2}}}{25}\text{=1}$ .