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(r – 1)² = –4(y – 1) y 9a² + 16(y – 2)² = 144 3 -2 1 0 2 5 -2 -3

(r – 1)² = –4(y – 1) y 9a² + 16(y – 2)² = 144 3 -2 1 0 2 5 -2 -3

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question

Consider the shaded region shown in the figure attached in the corresponding images, that involves the curves of equations:

(x − 1)2 = −4(y − 1)  and  9x2 + 16(y − 2)2 = 144

The area of ​​the shaded region roughly corresponds to:

Answers:

A) 6.0034 ul2

B) 19.6434 ul2

C) 19.6292 ul2

D) 2.8676 ul2

(r – 1)? = -4(y – 1) 9r? + 16(y – 2)² = 144 4 3 2 1 -2 - 4 -2 -3
Transcribed Image Text:(r – 1)? = -4(y – 1) 9r? + 16(y – 2)² = 144 4 3 2 1 -2 - 4 -2 -3
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