Chapter 10, Problem 18RE

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 18RE

Thevertex is $\left(\text{h,k}\right)\text{=}\left(\text{2,-1}\right)$ , Focus is $\left(\text{h-a, k}\right)\text{=}\left(2-\frac{3}{16}\text{,-1}\right)=\left(\frac{29}{16},-1\right)$ and Directrix is $x=h+a\Rightarrow x=2+\frac{3}{16}\Rightarrow \frac{35}{16}$ .

Explanation of Solution

Given:

  ${\text{4y}}^{\text{2}}\text{+3x-16y+19=0}$ .

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:

  ${\text{4y}}^{\text{2}}\text{+3x-16y+19=0}$ .

  ${\text{4y}}^{\text{2}}\text{-16y+16+3x=3=0}$ .

  $\text{4}\left(y^{\text{2}}\text{-4y+4}\right)\text{+3}\left(x+1\right)\text{=0}$ .

  $\Rightarrow \text{4}\left(y^{\text{2}}\text{-4y+4}\right)\text{=-3}\left(x+1\right).$

  $\Rightarrow {\left(y-2\right)}^2\text{=-}\frac{\text{3}}{4}\left(x+1\right).$

Here the equation is in the form of $\Rightarrow {\left(\text{x-h}\right)}^{\text{2}}\text{=-4a}\left(\text{y-k}\right)\text{.}$ where $\text{h=2,a=}\frac{\text{3}}{16}\text{ and k=-1}$ ,it’s parabola.

The vertex is $\left(\text{h,k}\right)\text{=}\left(\text{2,-1}\right)$ .

Focus is $\left(\text{h-a, k}\right)\text{=}\left(2-\frac{3}{16}\text{,-1}\right)=\left(\frac{29}{16},-1\right)$ .

Directrix is $x=h+a\Rightarrow x=2+\frac{3}{16}\Rightarrow \frac{35}{16}$ .

Hence, vertex is $\left(\text{h,k}\right)\text{=}\left(\text{2,-1}\right)$ , Focus is $\left(\text{h-a, k}\right)\text{=}\left(2-\frac{3}{16}\text{,-1}\right)=\left(\frac{29}{16},-1\right)$ and Directrix is $x=h+a\Rightarrow x=2+\frac{3}{16}\Rightarrow \frac{35}{16}$ .