Chapter 10, Problem 74RE

To determine

Tofind:the equation of the ellipse in which the vertices of the ellipseis equal to the foci of this hyperbola.

Answer to Problem 74RE

Theequation of ellipse is $\frac{{\text{x}}^{\text{2}}}{20}+\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

Explanation of Solution

Given:

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

Concept used:

Formula for ellipse:

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

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 foci are $\left(\text{±c,0}\right)$ where ${\text{c}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{-b}}^{\text{2}}$ .

Formula for hyperbola forming open left and right:

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

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

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

Calculation:

According to the given equation:

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

The equation is of hyperbola:

  $\frac{{\text{x}}^{\text{2}}}{16}-\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

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

  $a=4,\text{ b=2, c=}\sqrt{4^2+2^2}=2\sqrt{5}.$

Vertices of hyperbola= $\left(4,0\right)$ and $\left(-4,0\right)$ .

Foci of hyper bola= $\left(2\sqrt{5},0\right)$ and $\left(-2\sqrt{5},0\right)$ .

vertices of ellipse= $\left(2\sqrt{5},0\right)$ and $\left(-2\sqrt{5},0\right)$ .

Foci of ellipse= $\left(4,0\right)$ and $\left(-4,0\right)$ .

Vertices of ellipse= $\left(\text{a,0}\right)\text{ and }\left(\text{-a,0}\right)$ .

So, $\text{a=2}\sqrt{\text{5}}$ .

Now x-axis is major axis:

  $\text{c=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}$ .

  $\text{c=4}$ .

  ${\text{4}}^{\text{2}}{\text{=20-b}}^{\text{2}}$ .

  ${\text{b}}^{\text{2}}\text{=}\sqrt{\text{4}}\text{=2}$ .

So, equation of ellipse is

  $\frac{{\text{x}}^{\text{2}}}{20}+\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

Hence, the equation of ellipse is $\frac{{\text{x}}^{\text{2}}}{20}+\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .