Chapter 10, Problem 22RE

To determine

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

Answer to Problem 22RE

The equation of the conic described is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{25}}\text{=1}$ .

Explanation of Solution

Given:

Ellipse, center at $\left(0,0\right)$ focus at $\left(0,3\right)$ vertex at $\left(0,5\right)$ .

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:

Ellipse, center at $\left(0,0\right)$ focus at $\left(0,3\right)$ vertex at $\left(0,5\right)$ .

Centre= $\left(0,0\right)$ thus $\text{h,k}=0$ , foci and vertices now x-coordinate $0$ ,

Thus, equation is in the form:

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

$\text{focus=}\left(\text{0,c}\right)$ thus $\text{c=3}\text{.}$

  $\text{vertex=}\left(\text{0,a}\right)$

  $\text{thus a=5}$ .

  $\text{a=5,c=3,h=0,k=0 and b=}\sqrt{{\text{a}}^{\text{2}}{\text{-c}}^{\text{2}}}=\sqrt{25-9}=\sqrt{16}=4.$

The graph of the equation $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{25}}\text{=1}$ is

The graph here is of elliptical equation $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{25}}\text{=1}$ the graph here cut the x-axis 2time and y-axis 2times.

Hence, the equation of the conic described is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{25}}\text{=1}$ .