Undergrad Hyperbola and an ellipse to intersect orthogonally?

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The discussion focuses on the mathematical condition for a hyperbola and an ellipse to intersect orthogonally. The participants outline a systematic approach involving the equations of the curves, finding intersections, and calculating gradients. Specifically, they reference the hyperbola defined by the equation 2x² - 2y² = 1 and derive the eccentricity of the ellipse as 1/√2, leading to the relationship a² = 2b² for the ellipse. The steps provided are crucial for determining the equations of the curves that intersect orthogonally.

PREREQUISITES
  • Understanding of hyperbolas and ellipses in analytic geometry
  • Familiarity with gradient vectors and scalar products
  • Knowledge of eccentricity and its implications in conic sections
  • Ability to solve simultaneous equations
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  • Study the derivation of orthogonal intersections for conic sections
  • Learn about the properties of eccentricity in conic sections
  • Explore the method of Lagrange multipliers for constraints in optimization problems
  • Investigate the implications of gradient vectors in multi-variable calculus
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Mathematicians, students studying conic sections, and educators looking to deepen their understanding of the relationships between hyperbolas and ellipses.

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What is the condition for a hyperbola and an ellipse to intersect orthogonally?
I have a formula for orthogonal circles -> 2g1g2 + 2f1f2 - c1c2 = 0
 
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1) Write down the equations of the two curves.

2) Find the intersections by solving the two equations simultaneously.

3) Consider one of theese intersections, say (x0, y0).

4) Derive the first equation with respect to x and y. You get a vector u(x ,y) (the gradient).

5) Do the same with the second equation. Call the gradient v(x, y).

6) Evaluate u and v at the point (x0, y0) calculated earlier.

7) Calculate the scalar product u.v and impose it's zero.
 


Thanks for your response.
How do you apply rule 2 if you don't know the equation of one of the curves?
Consider this question-

An ellipse intersects the hyperbola 2x2 - 2y2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the co-ordinate axes, then find the equation of the ellipse.

Eccentricity of the ellipse is 1/√2

The relation between a and b of ellipse is a2 = 2b2

Now how do you proceed?
 


Well if what you say is correct (I didn't check it) then the two equations are:

x^2 - y^2 = 1/2

x^2 + 2y^2 = 2b^2

Try doing steps 2) - 7) with theese, you should get a constraint for the free parameter b.
 

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