Question

Prove Theorem 9.16 (b). That is, show that if $0<A<B$, the graph of the curve with equationrole="math" localid="1663152551040" $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ is an ellipse with foci role="math" localid="1663154443592" $\left(0,\pm \sqrt{B^2-A^2}\right)$.

Short answer

Hence, proved.

Step-by-step solution

Given information.

The equation of the ellipse is $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$

Plot the graph.

The graph of the ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$is,

The vertices of the parabola are $\left(0,\pm \sqrt{B^2-A^2}\right)$

And, from the graph it is observed that the foci of the parabola is the same $\left(0,\pm \sqrt{B^2-A^2}\right)$.

Hence, proved.