Chapter 10, Problem 10RE

To determine

Tofind:the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.

Answer to Problem 10RE

Thecenter= $\left(0,0\right)$ , vertices is $\left(0,\pm a\right)=\left(0,\pm 5\right)$ and Foci= $\left(0,\pm \text{c}\right)\text{=}\left(\text{0,}\pm \sqrt{5}\right)$ .

Explanation of Solution

Given:

  $9x^2+4y^2=36$ .

Concept used:

Formula for ellipse:

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

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:

  $9x^2+4y^2=36$ . Dividing by $36$

  $\frac{x^2}{4}+\frac{y^2}{9}=1$ .

Here the equation is in the form of $\frac{{\text{x}}^{\text{2}}}{{\text{a}}^{\text{2}}}+\frac{{\text{y}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ where $\text{a=5,b=1}$ ,it’s aellipse.

The center= $\left(0,0\right)$ The vertices is $\left(0,\pm a\right)=\left(0,\pm 5\right)$ .

Foci= $\left(0,\pm \text{c}\right)\text{=}\left(\text{0,}\pm \sqrt{5}\right)$ where $\text{c=}\sqrt{{\text{a}}^{\text{2}}{\text{-b}}^{\text{2}}}=\sqrt{9-4}=\sqrt{5}.$

Hence, center= $\left(0,0\right)$ , vertices is $\left(0,\pm a\right)=\left(0,\pm 5\right)$ and Foci= $\left(0,\pm \text{c}\right)\text{=}\left(\text{0,}\pm \sqrt{5}\right)$ .