Chapter 10, Problem 9RE

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 9RE

Thecenter= $\left(0,0\right)$ , vertices is $\left(\pm a,0\right)=\left(\pm \sqrt{2},0\right)$ , Foci= $\left(\pm \text{c,0}\right)\text{=}\left(\pm \sqrt{10}\text{,0}\right)$ and $\text{asymptotes:y=}\pm \sqrt{4}x=\pm 2x.$

Explanation of Solution

Given:

  $4x^2-y^2=8$ .

Concept used:

Formula for hyperbola forming open left and right.

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

Center= $\left(\text{h,k}\right)$ .

Vertices= $\left(\text{h+a,k}\right)$ and $\left(\text{h-a,k}\right)$ .

Slope of asymptotes: $\pm \frac{\text{b}}{a}$ .

Equation of asymptotes: $\text{y=k±}\frac{\text{b}}{\text{a}}\left(\text{x-h}\right)$ .

Formula for hyperbola opening up and down:

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

Center: $\left(\text{h,k}\right)$ .

Vertices: $\left(\text{h,k+b}\right)\text{ and }\left(\text{h,k-b}\right)$ .

Slope of asymptotes: $\text{±}\frac{\text{b}}{\text{a}}$ .

Calculation:

  $4x^2-y^2=8$ dividing by 8.

  $\frac{x^2}{2}-\frac{y^2}{8}=1$

Here the equation is in the form of $\frac{{\text{x}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{-}\frac{{\text{y}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ where $\text{a=}\sqrt{2}\text{,b=}\sqrt{8}$ ,it’s a hyperbola

The center= $\left(0,0\right)$ The vertices is $\left(\pm a,0\right)=\left(\pm \sqrt{2},0\right)$ .

Foci= $\left(\pm \text{c,0}\right)\text{=}\left(\pm \sqrt{10}\text{,0}\right)$ where $\text{c=}\sqrt{{\left(\sqrt{2}\right)}^{\text{2}}\text{+}{\left(\sqrt{8}\right)}^{\text{2}}}=\sqrt{10}.$

Asymptotes: $\text{y=}\pm \frac{b}{a}x\Rightarrow y=\pm \frac{\sqrt{8}}{\sqrt{2}}x$

  $\Rightarrow \text{y=}\pm \sqrt{4}x=\pm 2x.$

Hence, center= $\left(0,0\right)$ , vertices is $\left(\pm a,0\right)=\left(\pm \sqrt{2},0\right)$ , Foci= $\left(\pm \text{c,0}\right)\text{=}\left(\pm \sqrt{10}\text{,0}\right)$ and $\text{asymptotes:y=}\pm \sqrt{4}x=\pm 2x.$