Chapter 11.2, Problem 50E

To determine

To prove that all the ellipses represented by the given equation have the same foci, no matter what the value of $\text{k}$ is.

Explanation of Solution

Given information:

The given equation is

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

Formula used:

The following formula is used:

  ${\text{c}}^2={\text{a}}^2-{\text{b}}^2$

Proof:

In the given equation, $\frac{{\text{x}}^{\text{2}}}{\text{k}}+\frac{{\text{y}}^2}{\text{4+k}}=1;{\text{ a}}^2={\text{4+k and b}}^2=\text{k}$

The value of $\text{c}$ can be calculated by:

  $\begin{array}{l}{\text{c}}^2={\text{a}}^2-{\text{b}}^2\\ {\text{c}}^2=4+\text{k}-\text{k}\\ {\text{c}}^2=4\\ \text{c=}\pm \text{2}\end{array}$

The value of $\text{c}$ does not depend on the value of $\text{k}$ .

Hence, the equation will have the same foci, no matter what the value of $\text{k}$ .