Chapter 10, Problem 15RE

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 15RE

Thecenter= $\left(\text{h,k}\right)\text{=}\left(\text{2,1}\right)$ , vertices is $\left(\text{h±a,k}\right)\text{=}\left(\text{2±3,-1}\right)\text{=}\left(\text{5,-1}\right)\text{ and }\left(\text{-1,-1}\right)$ and Foci= $\left(2\pm \text{c,-1}\right)\text{=}\left(2\pm \sqrt{5}\text{,-1}\right)$ .

Explanation of Solution

Given:

  ${\text{4x}}^{\text{2}}{\text{+9y}}^{\text{2}}\text{-16x-18y=11}$ .

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:

  ${\text{4x}}^{\text{2}}{\text{+9y}}^{\text{2}}\text{-16x-18y=11}$ .

  ${\text{4x}}^{\text{2}}{\text{-16x+16+9y}}^{\text{2}}\text{-18y+9=11+9+16}$ .

  $\text{4}\left({\text{x}}^{\text{2}}\text{-4x+4}\right)\text{+9}\left({\text{y}}^{\text{2}}\text{-2y+1}\right)\text{=36}\text{.}$

  $\text{4}\frac{{\left(\text{x-2}\right)}^2}{36}\text{+9}\frac{{\left(\text{y-1}\right)}^2}{36}\text{=1}\text{.}$

  $\frac{{\left(\text{x-2}\right)}^2}{9}\text{+}\frac{{\left(\text{y-1}\right)}^2}{4}\text{=1}\text{.}$

Here the equation is in the form of $\frac{{\left(\text{x}-h\right)}^{\text{2}}}{{\text{a}}^{\text{2}}}+\frac{{\left(\text{y-k}\right)}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ where $\text{a=3,b=2}$ ,it’s an ellipse.

The center= $\left(\text{h,k}\right)\text{=}\left(\text{2,1}\right)$ The vertices is $\left(\text{h±a,k}\right)\text{=}\left(\text{2±3,-1}\right)\text{=}\left(\text{5,-1}\right)\text{ and }\left(\text{-1,-1}\right)$ .

Foci= $\left(2\pm \text{c,-1}\right)\text{=}\left(2\pm \sqrt{5}\text{,-1}\right)$ where $\text{c=}\sqrt{{\text{a}}^{\text{2}}{\text{-b}}^{\text{2}}}=\sqrt{5}.$

Hence, center= $\left(\text{h,k}\right)\text{=}\left(\text{2,1}\right)$ , vertices is $\left(\text{h±a,k}\right)\text{=}\left(\text{2±3,-1}\right)\text{=}\left(\text{5,-1}\right)\text{ and }\left(\text{-1,-1}\right)$ and Foci= $\left(2\pm \text{c,-1}\right)\text{=}\left(2\pm \sqrt{5}\text{,-1}\right)$ .