Chapter 10, Problem 20RE

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 20RE

Thecenter= $\left(\text{h,k}\right)\text{=}\left(\text{1,-1}\right)$ ,Foci= $\left(\text{h}\pm \text{c,k}\right)\text{=}\left(\text{1}\pm \sqrt{2}\text{,-1}\right)$ and $\text{Asymptotes:y=x-2 or y=-x}\text{.}$

Explanation of Solution

Given:

  ${\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{-2x-2y=1}$ .

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:

  ${\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{-2x-2y=1}$ .

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

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

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

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

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

Asymptotes: $y-k=\pm 1\left(x-1\right)$

  $\Rightarrow y=x-1-1or\text{ y=-x+1-1}$

  $\text{Asymptotes:y=x-2 or y=-x}\text{.}$

Hence, center= $\left(\text{h,k}\right)\text{=}\left(\text{1,-1}\right)$ ,Foci= $\left(\text{h}\pm \text{c,k}\right)\text{=}\left(\text{1}\pm \sqrt{2}\text{,-1}\right)$ and $\text{Asymptotes:y=x-2 or y=-x}\text{.}$