Chapter 10, Problem 14RE

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 14RE

Thecenter= $\left(\text{h,k}\right)\text{=}\left(\text{-1,2}\right)$ , vertices is $\left(\text{h,k}\pm \text{a}\right)\text{=}\left(\text{-1,2}\pm \text{2}\right)\text{=}\left(\text{-1,0}\right)\text{ and }\left(\text{-1,4}\right)$ and Foci= $\left(\text{h,k}\pm \text{c}\right)\text{=}\left(\text{-1,2}\pm \sqrt{3}\right)$ .

Explanation of Solution

Given:

  ${\text{4x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{+8x-4y+4=0}$ .

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:

  ${\text{4x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{+8x-4y+4=0}$ .

Adding $4$ on both side:

  ${\text{4x}}^{\text{2}}{\text{+8x+4+y}}^{\text{2}}\text{+4y+4=4}$ .

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

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

Or

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

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 $\text{a=2,b=1}$ ,it’s an ellipse.

The center= $\left(\text{h,k}\right)\text{=}\left(\text{-1,2}\right)$ The vertices is $\left(\text{h,k}\pm \text{a}\right)\text{=}\left(\text{-1,2}\pm \text{2}\right)\text{=}\left(\text{-1,0}\right)\text{ and }\left(\text{-1,4}\right)$ .

Foci= $\left(\text{h,k}\pm \text{c}\right)\text{=}\left(\text{-1,2}\pm \sqrt{3}\right)$ where $\text{c=}\sqrt{{\text{a}}^{\text{2}}{\text{-b}}^{\text{2}}}=\sqrt{3}.$

Hence, center= $\left(\text{h,k}\right)\text{=}\left(\text{-1,2}\right)$ , vertices is $\left(\text{h,k}\pm \text{a}\right)\text{=}\left(\text{-1,2}\pm \text{2}\right)\text{=}\left(\text{-1,0}\right)\text{ and }\left(\text{-1,4}\right)$ and Foci= $\left(\text{h,k}\pm \text{c}\right)\text{=}\left(\text{-1,2}\pm \sqrt{3}\right)$ .