Question

Find an equation of the hyperbola whose foci are the vertices of the ellipse $4x^2+9y^2=36$and whose vertices are the foci of this ellipse.

Short answer

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

Step-by-step solution

Step 1. Given Information

Given to determine the equation of the hyperbola whose foci are the vertices of the ellipse $4x^2+9y^2=36$and whose vertices are the foci of this ellipse.

Step 2. Analyzing the ellipse

Rewriting the given equation of ellipse:

$$\begin{gathered} \frac{4x^2}{36}+\frac{9y^2}{36}=\frac{36}{36} \\ \frac{x^2}{9}+\frac{y^2}{4}=1 \end{gathered}$$

Here $a=3,b=2$

Hence the vertices of the parabola are $\left(-3,0\right),\left(3,0\right)$

The foci of this ellipse lie at $\left(-c,0\right),\left(c,0\right)$where $c^2=a^2-b^2$.

Plugging the values:

localid="1648614483350" $$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=9-4 \\ c=\pm \sqrt{5} \end{gathered}$$

Hence the foci of the ellipse are$\left(-\sqrt{5},0\right),\left(\sqrt{5},0\right)$

Step 3. Equation of hyperbola

So the vertices of the hyperbola are $\left(-\sqrt{5},0\right),\left(\sqrt{5},0\right)$and the foci are $\left(-3,0\right),\left(3,0\right)$.

Here, role="math" localid="1648122490904" $a=\sqrt{5},c=3$

For a hyperbola,

$$\begin{gathered} c^2=a^2+b^2 \\ 9=5+b^2 \\ b=2 \end{gathered}$$

Hence the equation of the hyperbola is given by:

$$\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \frac{x^2}{5}-\frac{y^2}{4}=1 \end{gathered}$$

Step 4. Conclusion

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