Chapter 10, Problem 25RE

To determine

Tofind:the equation of conic using its given description and graph the equation.

Answer to Problem 25RE

The parabolic equation is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{7}}\text{=1}$ .

Explanation of Solution

Given:

Ellipse, foci at $\left(-3,0\right)$ and $\left(3,0\right)$ vertex is at $\left(\text{h,k}\right)$ = $\left(4,0\right)$ .

Concept used:

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

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:

According to the given:

Ellipse, foci at $\left(-3,0\right)$ and $\left(3,0\right)$ vertex is at $\left(\text{h,k}\right)$ = $\left(4,0\right)$ .

It resembled ellipse in the form of:

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

Here foci = $\left(\text{h±c,k}\right)$ .

Comparing with given data:

Thus $\text{k=0,h=0,c=3}$ .

  $\text{Vertex=}\left(\text{h+a,k}\right)$

Thus $\text{a=4,b=}\sqrt{{\text{a}}^{\text{2}}{\text{-c}}^{\text{2}}}=\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt{7.}$

  $\text{a=4,b=}\sqrt{\text{7}}$ .

Now equation will be:

  $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{7}}\text{=1}$

The equation can be in shown in graph as:

The graph of parabolic equation $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{7}}\text{=1}$ .

Hence, parabolic equation is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{7}}\text{=1}$ .