Confocal ellipse and hyperbola

  • Thread starter Thread starter Ananya0107
  • Start date Start date
  • Tags Tags
    Ellipse Hyperbola
AI Thread Summary
A hyperbola that passes through the foci of the ellipse defined by the equation x^2/25 + y^2/16 = 1 has its transverse and conjugate axes aligned with the ellipse's major and minor axes. The product of the eccentricities of the hyperbola and ellipse equals 1, leading to the hyperbola's eccentricity being 5/3. The derived equation of the hyperbola is x^2/9 - y^2/9 = 1, with its foci located at ±5,0. The discussion emphasizes the importance of correctly applying given information in coordinate geometry, noting the complexity of related problems. Understanding the relationship between confocal ellipses and hyperbolas is crucial in solving these types of equations.
Ananya0107
Messages
20
Reaction score
2
If a hyperbola passes through the focii of the ellipse x^2/25 +y^2/16 =1 and its transverse and conjugate axes coincide respectively with major and minor axes of the ellipse, and if the product of eccentricities of hyperbola and ellipse is 1, find the equation and focus of the hyperbola
 
Mathematics news on Phys.org
There is a very important property regarding confocal ellipse and hyperbola.
"When ellipse and hyperbola are confocal, then they are orthogonal curves"
 
  • Like
Likes Ananya0107
But I don't think you need that property here. I was thinking about finding the differential equation for the ellipse and then substituting -dx/dy for dy/dx and then finding the curve equation for hyperbola by integrating. You can try it.
 
A second method is to find e for ellipse, then find focus, then the hyperbola equation by using the information of product of eccentricities.
 
Actua
AdityaDev said:
But I don't think you need that property here. I was thinking about finding the differential equation for the ellipse and then substituting -dx/dy for dy/dx and then finding the curve equation for hyperbola by integrating. You can try it.
actually I was thinking too much ...:sorry: 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 ±5,0
 
Coordinate geometry is one topic which tests how much properly you can use the given information. But there are very difficult questions in this topic.
 
True..
 
Back
Top