actually I was thinking too much ...AdityaDev said:
Eccentricity of the hyperbola = 5/3 from the question , and it passes through (±3, 0) , its equation therefore is x^2/9 - y^2/b^2 =1 where 1+ b^2/9 = 25/9 therefore equation of hyperbola is x^2/9 - y^2/9 = 1 and its focii are ±5,0A confocal ellipse and hyperbola refer to a set of curves that share the same foci. This means that they have a common point or focus that all of the curves pass through.
Confocal ellipses and hyperbolas are related because they both have the same set of foci. They differ in their shape, with an ellipse being a closed curve while a hyperbola is an open curve.
Confocal ellipses and hyperbolas have various applications in fields such as optics, astronomy, and engineering. They can be used to describe the path of light rays, orbits of celestial bodies, and the shape of reflectors and antennas.
Confocal ellipses and hyperbolas can be represented using their respective equations: for an ellipse, it is (x/a)^2 + (y/b)^2 = 1, and for a hyperbola, it is (x/a)^2 - (y/b)^2 = 1. The parameters a and b represent the lengths of the semi-major and semi-minor axes of the curves.
The main difference between a confocal ellipse and a non-confocal ellipse is that a confocal ellipse has a set of foci that all of the curves share, while a non-confocal ellipse does not. In a non-confocal ellipse, the foci are located at different points, and the curve does not pass through them.