Question 13

From the ellipse, I get a²=18, b²=12, c²=6, giving eccentricity 1/√3 and foci at (±√6, 0). Using the hyperbola's eccentricity of √3 with the same foci, I find A²=2 and B²=4, so the hyperbola is x²/2 − y²/4 = 1. Substituting the parabola x²=√5 y into the hyperbola equation gives y²−2√5y+4=0, so y=√5±1 and x²=5±√5, picking the positive x roots for the first quadrant. Computing the squared distance between these two points using x1²+x2²=10 and the product x1x2=√20, I get d²=14−4√5, so with a=14 and b=−4, a−b=18. **Setting up** Ellipse: a² = 18, b² = 12 ⇒ c² = 6, e = 1/√3, foci (±√6, 0). Hyperbola: e' = √3, c = √6 ⇒ A = c/e' = √2, A² = 2, B² = c² − A² = 4. So H: x²/2 − y²/4 = 1. **Intersection with √5 y = x²** Substitute x² = √5 y: √5 y/2 − y²/4 = 1 ⇒ y² − 2√5 y + 4 = 0 ⇒ y = √5 ± 1. Then x² = 5 ± √5, with x > 0 in the first quadrant. **Distance** d² = (x₁ − x₂)² + (y₁ − y₂)² = (x₁² + x₂² − 2x₁x₂) + 4 = 10 − 2√(25 − 5) + 4 = 14 − 4√5. So a = 14, b = −4, giving a − b = 18.

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For the ellipse, a²=18, b²=12, giving c²=6 and eccentricity 1/√3. Since the hyperbola shares foci with the ellipse and has eccentricity √3, I get c²=6 = a_H²·3, so a_H²=2 and b_H²=4, giving the hyperbola equation x²/2 - y²/4 = 1. Substituting the parabola x²=√5y into the hyperbola gives y² - 2√5y + 4 = 0, so y = √5+1 or √5-1. Both are positive, matching the first-quadrant condition, so x1² = 5+√5 and x2² = 5-√5. Computing the squared distance between the two points: (y1-y2)²=4, and (x1-x2)² = 10 - 2√20 = 10-4√5, so overall d² = 10. So d² = 10-4√5+4 = 14-4√5, giving a=14, b=-4, and a-b = 18. **Solution Summary:** - Ellipse: a²=18, b²=12, e²=1/3 - Hyperbola: e_H=√3, foci same (c²=6), so a_H²=2, b_H²=4 - H: x²/2 − y²/4 = 1 Intersecting with x² = √5 y: (√5 y)/2 − y²/4 = 1 → y² − 2√5y + 4 = 0 → y = √5 ± 1 Corresponding x² = √5y → x² = 5+√5 or 5−√5 d² = (x₁−x₂)² + (y₁−y₂)² = [(5+√5)+(5−√5) − 2√((5+√5)(5−√5))] + (2)² = 10 − 2√20 + 4 = 14 − 4√5 So a = 14, b = −4, giving a − b = 18. ```json