Chapter 10, Problem 33RE

To determine

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

Answer to Problem 33RE

The elliptical equation is $\frac{{\left(\text{x+1}\right)}^{\text{2}}}{9}\text{+}\frac{{\left(\text{y-2}\right)}^{\text{2}}}{8}\text{=1}$ .

Explanation of Solution

Given:

Ellipse, center at $\left(-1,2\right)$ .

Focus at $\left(0,2\right)$ and vertex $\left(2,0\right)$ .

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:

Centre at $\left(-1,2\right)$ ; $a=3$ ; $c=4$ .

  ${\text{c}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{-b}}^{\text{2}}$ .

  ${\text{16=9-b}}^{\text{2}}$ .

  ${\text{-7=b}}^{\text{2}}$ .

It resembles ellipse in the form:

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

here, $\text{h=-1,k=2}$

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

focus= $\left(\text{h+c, k}\right)\text{=}\left(\text{3,2}\right)$ .

The equation of the ellipse is:

  $\frac{{\left(\text{x+1}\right)}^{\text{2}}}{3^{\text{2}}}+\frac{{\left(\text{y-2}\right)}^{\text{2}}}{\left(7\right)}\text{=1}$ .

The equation can be in shown in graph as:

The graph of elliptical equation $\frac{{\left(\text{x+1}\right)}^{\text{2}}}{3^{\text{2}}}+\frac{{\left(\text{y-2}\right)}^{\text{2}}}{\left(7\right)}\text{=1}$ .

  

From the graph:

At $x=-2.964;\text{ y=0}$ , $x=0.964;\text{ y=0}$ .

At $x=0;\text{ y=4}\text{.494}$ , $x=0;\text{ y=-0}\text{.494}$ .

Hence, the equation is ellipse.