(x^2)/(a^2) + (y^2)/(b^2) = 1


(x^2)/(a^2-2b^2) - (y^2)/(b^2) = 1 where a > b(sqrt(2))

1. Show that the ellipse and the hyperbola are confocal

2. At how many points do these curves intersect?

3. Let e and E be the eccentricities of this ellipse and hyperbola, respectively. Also, let C^2 = a^2 –b^2 . Show that the coordinates of the point in the first quadrant where the curves intersect are given by x= (C)/(eE) and y=(b^2)/(C)
asked May 13 in Calculus Answers by anonymous

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1 Answer

    1    For the ellipse the focus is given by (f,0) where f=√a²-b². For the hyperbola the focus is (f,0) where f=√(a²-2b²)+b²=√a²-b², as for the ellipse, so both conics are confocal.

    2    If we write the equation for the ellipse y²/b²=1-x²/a² and for the hyperbola y²/b²=x²/(a²-2b²)-1, then we can equate the expressions in x: 1-x²/a²=x²/(a²-2b²)-1. Multiply through by a²(a²-2b²) we get: 2a²(a²-2b²)-x²(a²-2b²)=a²x², from which x²=a²(a²-2b²)/(a²-b²). So x²/a²=(a²-2b²)/(a²-b²) and y²/b²=1-(a²-2b²)/(a²-b²)=b²/(a²-b²) and y=b²/√(a²-b²) [which can also be written b²/C if C²=a²-b²].

There are four points of intersection:

(a√((a²-2b²)/(a²-b²)), b²/√(a²-b²));

(a√((a²-2b²)/(a²-b²)), -b²/√(a²-b²));

(-a√((a²-2b²)/(a²-b²)), b²/√(a²-b²));

(-a√((a²-2b²)/(a²-b²)), -b²/√(a²-b²)).

    3    The first quadrant intercept is at x=a√((a²-2b²)/(a²-b²))=a√(a²-2b²)/C, y=b²/C.

b²=a²(1-e²) for the ellipse, so e=√(a²-b²)/a, and E=√((a²-2b²+b²)/(a²-2b²))=√((a²-b²)/(a²-2b²)) for the hyperbola. Therefore 1/E=√((a²-2b²)/(a²-b²)) and 1/e=a/√(a²-b²).

So 1/eE=a√(a²-2b²)/C²=x/C, therefore x=C/eE. The first quadrant intercept is therefore at x=C/eE and y=b²/C, QED.



answered May 14 by Rod Top Rated User (581,240 points)

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