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Orthogonal Conics (Posted on 2011-12-10)

An ellipse and hyperbola have the same foci.

Prove that they are orthogonal.

See The Solution Submitted by Bractals

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Solution

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The ellipse         b²x² + a²y² = a²b²   has foci at (+/- sqrt(a² – b²), 0)

The hyperbola    B²x² – A²y² = A²B² has foci at (+/- sqrt(A² + B²), 0)

So, for coincident foci:    a² – b² = A² + B², giving a² – A² = b² + B²    (1)

For intersection at (x₀, y₀):

            b²x₀² + a²y₀² = a²b²         and       B²x₀² – A²y₀² = A²B²

Eliminating y₀ and x₀, respectively, from these two equations gives:

x₀²(b²A² + a²B²) = a²A²(b² + B²) ,   y₀²(b²A² + a²B²) = b²B²(a² – A²)

Using equation (1), it follows that:            x₀²b²B² = y₀²a²A² (2)

The gradients of the ellipse and hyperbola are, respectively:

            y' = -b²x/(a²y)   and       y' = B²x/(A²y)

so the product of their gradients at the point of intersection is

            -b²B²x₀² / a²A²y₀² = -1               from equation (2)

So the ellipse and hyperbola are orthogonal.

Edited on December 12, 2011, 10:15 pm

Posted by Harry on 2011-12-12 22:14:17

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