Question

Consider the ellipse \(E\) given by \[ \frac{x^2}{18} + \frac{y^2}{12} = 1. \] Let \(H\) be the hyperbola whose eccentricity is the reciprocal of the eccentricity of \(E\) and whose foci are the same as that of \(E\). Let \(P\) and \(Q\) be the points of intersection of \(H\) and the parabola \[ \sqrt{5}\,y = x^2 \] in the first quadrant. Let \(d\) be the distance between \(P\) and \(Q\). If \(a\) and \(b\) are the integers such that \[ d^2 = a + b\sqrt{5}, \] then the value of \[ a - b \] is:

Hint:

Confocal conics share the term $ae$. In these problems, always find the distance of the focus from the center first, as it links the ellipse and hyperbola directly.

Answer 1

Correct Answer: 18

Step 1: Understanding the Question:
This problem integrates multiple conic sections. We need to find the equation of a hyperbola $H$ derived from an ellipse $E$, find its intersection points with a parabola, and then calculate the squared distance between these points.

Step 2: Key Formula or Approach:

• Eccentricity of ellipse $e_E^2 = 1 - b^2/a^2$.

• For hyperbola $e_H = 1/e_E$. Foci are $(\pm ae_E, 0)$.

• Distance formula $d^2 = (x_1-x_2)^2 + (y_1-y_2)^2$.

Step 3: Detailed Explanation:

• For ellipse $E: a^2=18, b^2=12 \implies e_E^2 = 1 - \frac{12}{18} = \frac{1}{3} \implies e_E = \frac{1}{\sqrt{3}}$.

• Eccentricity of hyperbola $e_H = \sqrt{3}$.

• Foci of $E$: $(\pm \sqrt{18 \cdot 1/3}, 0) = (\pm \sqrt{6}, 0)$. These are also foci of $H$.

• For $H$: $Ae_H = \sqrt{6} \implies A\sqrt{3} = \sqrt{6} \implies A = \sqrt{2}$. $A^2 = 2$.
$B^2 = A^2(e_H^2 - 1) = 2(3-1) = 4$.
Equation $H: \frac{x^2}{2} - \frac{y^2}{4} = 1$.

• Intersection with parabola $x^2 = \sqrt{5}y$:
$\frac{\sqrt{5}y}{2} - \frac{y^2}{4} = 1 \implies 2\sqrt{5}y - y^2 = 4 \implies y^2 - 2\sqrt{5}y + 4 = 0$.
Roots: $y = \frac{2\sqrt{5} \pm \sqrt{20 - 16}}{2} = \sqrt{5} \pm 1$.
$y_1 = \sqrt{5}+1$ and $y_2 = \sqrt{5}-1$. Both are positive (first quadrant).

• Corresponding $x$ values:
$x_1^2 = \sqrt{5}(\sqrt{5}+1) = 5+\sqrt{5}$.
$x_2^2 = \sqrt{5}(\sqrt{5}-1) = 5-\sqrt{5}$.

• Distance $d^2 = (x_1-x_2)^2 + (y_1-y_2)^2$.
$(y_1-y_2)^2 = (2)^2 = 4$.
$(x_1-x_2)^2 = x_1^2 + x_2^2 - 2x_1x_2 = (10) - 2\sqrt{25-5} = 10 - 2\sqrt{20} = 10 - 4\sqrt{5}$.

• $d^2 = 10 - 4\sqrt{5} + 4 = 14 - 4\sqrt{5}$.
Comparing with $a+b\sqrt{5}$, $a=14$ and $b=-4$.
Value $a-b = 14 - (-4) = 18$.

Step 4: Final Answer:
The value of $a-b$ is 18.