Let e_1 and e_2 be the eccentricities of the ellipse x^2b^2+ y^225=1 and the hyperbola x^216- y^2b^2=1 , respectively. If b

Question

Let e_1 and e_2 be the eccentricities of the ellipse x^2b^2+ y^225=1 and the hyperbola x^216- y^2b^2=1 , respectively. If b<5 and e_1e_2=1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

Accepted Answer

Correct answer: B. (b) Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci. Find e_1 for the Ellipse The equation of the ellipse is: x^2b^2+ y^225=1 The eccentricity e_1 is given by: e_1^2=1- b^225 Find e_2 for the Hyperbola The equation of the hyperbola is: x^216- y^2b^2=1 The eccentricity e_2 is given by: e_2^2=1+ b^216 Using the Product e_1e_2=1 Given: e_1e_2=1 Thus: (1- b^225)(1+ b^216)=1 Expanding gives: 1+ b^216- b^225- b^4400=1 Simplifying: 9b^2400= b^4400 Thus: b^2=9 Determine Eccentricities e_1 and e_2 Substitute b^2=9 : For the ellipse: e_1^2=1- 925= 1625e_1= 45 For the hyperbola: e_2= 54 Find the Eccentricity of the New Ellipse The new ellipse's equation is: x^225+ y^216=1 The eccentricity e is: e= 1- 1625= 925= 35 Thus, the eccentricity of the ellipse that passes through all four foci is 35.

Solution

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