Chapter 10, Problem 11RE

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 11RE

Thevertex= $\left(\text{h,k}\right)\text{=}\left(\text{2,-2}\right)$ ,Focus= $\left(\text{h,k+a}\right)\text{=}\left(\text{2,-2+}\frac{1}{2}\right)=\left(2,-\frac{3}{2}\right)$ and Directrix: $\text{y=k-a}\Rightarrow y=-2-\frac{1}{2}.$

Explanation of Solution

Given:

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

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:

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

Adding 4 on both side:

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

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

Which is in the form ${\left(\text{x-h}\right)}^{\text{2}}\text{=4a}\left(\text{y-k}\right)$ .

Where $\text{h=2 and k=-2 and a=}\frac{1}{2}.$

Thus, the parabola vertex= $\left(\text{h,k}\right)\text{=}\left(\text{2,-2}\right)$ .

Focus= $\left(\text{h,k+a}\right)\text{=}\left(\text{2,-2+}\frac{1}{2}\right)=\left(2,-\frac{3}{2}\right)$ .

Directrix: $\text{y=k-a}\Rightarrow y=-2-\frac{1}{2}.$

  $y=-\frac{5}{2}$ .

Hence, vertex= $\left(\text{h,k}\right)\text{=}\left(\text{2,-2}\right)$ ,Focus= $\left(\text{h,k+a}\right)\text{=}\left(\text{2,-2+}\frac{1}{2}\right)=\left(2,-\frac{3}{2}\right)$ and Directrix: $\text{y=k-a}\Rightarrow y=-2-\frac{1}{2}.$