Chapter 10, Problem 73RE

To determine

Tofind:the equation of the hyperbola in which the vertices of the hyperbola is equal to the fociof this ellipse.

Answer to Problem 73RE

The equation of hyperbola is $\frac{{\text{x}}^{\text{2}}}{5}\text{-}\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

Explanation of Solution

Given:

  ${\text{4x}}^{\text{2}}{\text{+9y}}^{\text{2}}\text{=36}$ .

Vertices of the hyperbola is equal to the foci of the ellipse.

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:

  ${\text{4x}}^{\text{2}}{\text{+9y}}^{\text{2}}\text{=36}$ .

The equation is of ellipse.

  $\frac{{\text{x}}^{\text{2}}}{9}\text{+}\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

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

Vertices= $\left(3,0\right)$ and $\left(-3,0\right)$ .

Foci= $\left(\sqrt{{\text{a}}^{\text{2}}{\text{-b}}^{\text{2}}}\text{,0}\right)$ and $\left(-\sqrt{{\text{a}}^{\text{2}}{\text{-b}}^{\text{2}}}\text{,0}\right)$ .

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

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

Foci of hyperbola= $\left(3,0\right)$ and $\left(-3,0\right)$ .

Let equation of hyperbola:

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

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

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

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

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

  $\text{3=}\sqrt{{\text{5+b}}^{\text{2}}}$ .

  ${\text{b}}^{\text{2}}=\sqrt{9-5}=2$ .

So, equation of hyperbola is $\frac{{\text{x}}^{\text{2}}}{5}\text{-}\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .

Hence, the equation of hyperbola is $\frac{{\text{x}}^{\text{2}}}{5}\text{-}\frac{{\text{y}}^{\text{2}}}{4}\text{=1}$ .