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in Mathematics by (53.1k points)

Let a hyperbola pass through the focus of the ellipse

x2/25 + y2/16 = 1

The transeverse and conjugate axes of this hyperbola coincide with the major and minor axes of the ellipse. Also the product of the eccentricities of the ellipse and hyperbola is 1. Then

(A)  the equation of the hyperbola is x2/9 - y2/16 = 1

(B)  the equation of the hyperbola is x2/9 - y2/25 = 1

(C)  the focus of the hyperbola is (5, 0)

(D)  the vertex of hyperbola is (5√3,0)

1 Answer

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by (53.3k points)
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Best answer

Correct option  (a),(c)

Explanation :

The eccentricity e of the ellipse is given by

16 = 25(1 - e2)

which gives that

e = 3/5

Hence, the eccentricity of the hyperbola is 5/3. Let

x22 - y2/β2 = 1

be the hyperbola. Now,

β2 = α(25/9 - 1) = 16α2/9

This implies that the equation is

x22 - 9y2/16α2 = 1

Also the hyperbola passes through the focus (3, 0) of the ellipse. This implies that

Hence, the equation of the hyperbola is

x2/9 - y2/16 = 1

One vertex is (3, 0) and the focus is

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