# Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 457: 43

The points are: $(5,0); (-5,0); (0,-4); (0,4)$

#### Work Step by Step

We solve for y: $2\sqrt{9+y^2} = 10 \\ \sqrt{9+y^2} = 5 \\ y = \pm4$ We solve for the x values of the y-intercepts as well: $10 = \sqrt{(3+x)^2} + \sqrt{(3-x)^2} \\ 2x = \pm 10 \\ x = \pm 5$ To draw an ellipse, use curvy lines to connect the points found.

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