Hyperbola

Concept:The hyperbola has the same foci as the ellipse, and its eccentricity is the reciprocal of the ellipse's eccentricity. The distance between the two intersection points of the hyperbola and the given parabola in the first quadrant is found, then expressed in the form $a + b\sqrt{5}$ to compute $a-b$.Explanation:The ellipse is $\frac{x^2}{18}+\frac{y^2}{12}=1$, so $a^2=18$, $b^2=12$.Ellipse eccentricity: $e_E = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{12}{18}} = \frac{1}{\sqrt{3}}$.Distance of foci from centre: $a e_E = \sqrt{18}\cdot \frac{1}{\sqrt{3}} = \sqrt{6}$.Hence foci are $(\pm\sqrt{6},0)$.Hyperbola eccentricity: $e_H = \frac{1}{e_E} = \sqrt{3}$.Let hyperbola be $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$. Foci: $(\pm A e_H,0) = (\pm\sqrt{6},0)$.Thus $A e_H = \sqrt{6} \Rightarrow A = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$, so $A^2=2$.For hyperbola, $e_H^2 = 1+\frac{B^2}{A^2} \Rightarrow 3 = 1+\frac{B^2}{2} \Rightarrow B^2=4$.Hyperbola equation: $\frac{x^2}{2}-\frac{y^2}{4}=1$.Parabola: $\sqrt{5}\,y = x^2 \Rightarrow x^2 = \sqrt{5}y$.Substitute into hyperbola: $\frac{\sqrt{5}y}{2} - \frac{y^2}{4}=1$. Multiply by 4: $2\sqrt{5}y - y^2 = 4$.Rearrange: $y^2 - 2\sqrt{5}y + 4 =0$. Solve: $y = \frac{2\sqrt{5} \pm \sqrt{20-16}}{2} = \sqrt{5}\pm1$.Thus $y_1 = \sqrt{5}+1$, $y_2 = \sqrt{5}-1$.Corresponding $x_1^2 = \sqrt{5}(\sqrt{5}+1)=5+\sqrt{5}$, $x_2^2 = \sqrt{5}(\sqrt{5}-1)=5-\sqrt{5}$.Since points are in first quadrant, $x_i>0$, so $x_1,x_2$ are positive. We need distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$.Compute $d^2 = (x_1-x_2)^2 + (y_1-y_2)^2$.$(y_1-y_2)^2 = [(\sqrt{5}+1)-(\sqrt{5}-1)]^2 = 2^2 =4$.$(x_1-x_2)^2 = x_1^2 + x_2^2 - 2x_1x_2$.$x_1^2 + x_2^2 = (5+\sqrt{5})+(5-\sqrt{5})=10$.$x_1x_2 = \sqrt{x_1^2 x_2^2} = \sqrt{(5+\sqrt{5})(5-\sqrt{5})} = \sqrt{25-5} = \sqrt{20}=2\sqrt{5}$.Hence $(x_1-x_2)^2 = 10 - 2(2\sqrt{5}) = 10 - 4\sqrt{5}$.Therefore $d^2 = (10-4\sqrt{5}) + 4 = 14 - 4\sqrt{5}$.So $a=14$, $b=-4$. Then $a-b = 14 - (-4) = 18$.Answer:18