Chapter 9: Q. 55 (page 775)

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" $(0, \pm \sqrt{B^{2} - A^{2}})$ .

Short Answer

Expert verified

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 $(0, \pm \sqrt{B^{2} - A^{2}})$

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

Hence, proved.

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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" $(0, \pm \sqrt{B^{2} - A^{2}})$ .

Short Answer

Step by step solution

Given information.

Plot the graph.

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Prove Theorem 9.16 (b). That is, show that if 0<A<B, the graph of the curve with equationrole="math" localid="1663152551040" x2A2+y2B2=1 is an ellipse with foci role="math" localid="1663154443592" 0,±B2-A2.

Short Answer

Step by step solution

Given information.

Plot the graph.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

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" $(0, \pm \sqrt{B^{2} - A^{2}})$ .