Chapter 10, Problem 12RE

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 12RE

Thevertex= $\left(\text{h,k}\right)\text{=}\left(\text{0,1}\right)$ ,Focus= $\left(\text{h+a,k}\right)\text{=}\left(\frac{1}{8}\text{,1}\right)$ and $\text{Directrix=-}\frac{\text{1}}{\text{8}}$ .

Explanation of Solution

Given:

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

Concept used:

The standard form is ${\left(\text{x-h}\right)}^{\text{2}}\text{=4p}\left(\text{y-k}\right)$ where the focus is $\left(\text{h,k+p}\right)$ and the directrix is $\text{y=k-p}$ . If the parabola is rotated so that its vertex is $\left(\text{h,k}\right)$ and its axis of symmetry is parallel to the x-axis, it has an equation of ${\left(\text{y-k}\right)}^{\text{2}}\text{=4p}\left(\text{x-h}\right)$ where the focus is $\left(\text{h+p,k}\right)$ and the directrix is $\text{x=h-p}$ .

Calculation:

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

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

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

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

Which is in the form of:

  $\Rightarrow {\left(\text{y-k}\right)}^{\text{2}}\text{=4a}\left(\text{x-h}\right)$

Where $k=1,\text{ h=0 and a=}\frac{1}{8}$ .

Thus, the parabola vertex= $\left(\text{h,k}\right)\text{=}\left(\text{0,1}\right)$ .

Focus= $\left(\text{h+a,k}\right)\text{=}\left(\frac{1}{8}\text{,1}\right)$ .

  $\Rightarrow x=h-a\Rightarrow x=0-\frac{1}{8}$ .

  $\text{Directrix=-}\frac{\text{1}}{\text{8}}$ .

Hence, vertex= $\left(\text{h,k}\right)\text{=}\left(\text{0,1}\right)$ ,Focus= $\left(\text{h+a,k}\right)\text{=}\left(\frac{1}{8}\text{,1}\right)$ and $\text{Directrix=-}\frac{\text{1}}{\text{8}}$ .