Conics — Ellipse, Hyperbola & Parabola Intersections
PYP_JEE_ADV_2026_P2
Grade None
Question:
Consider the ellipse $E$ given by $\dfrac{x^2}{18}+\dfrac{y^2}{12}=1$. 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 __________.
Step-by-Step Solution
Key Concept: The reciprocal eccentricity relation $e_H=1/e_E$ links the two conics. The distance formula requires careful treatment of the square root terms using the identity $(\sqrt{A}-\sqrt{B})^2=A+B-2\sqrt{AB}$.
**Step 1: Find eccentricities and hyperbola**
$E$: $e_E=\sqrt{1-12/18}=1/\sqrt{3}$. $H$: $e_H=\sqrt{3}$, foci at $(\pm\sqrt{6},0)$. $a_H=\sqrt{6}/\sqrt{3}=\sqrt{2}$, $b_H^2=6-2=4$. $H$: $\dfrac{x^2}{2}-\dfrac{y^2}{4}=1$.
**Step 2: Find intersection with parabola $x^2=\sqrt{5}y$**
$\frac{\sqrt{5}y}{2}-\frac{y^2}{4}=1 \Rightarrow y^2-2\sqrt{5}y+4=0 \Rightarrow y=\sqrt{5}\pm1$. First quadrant: $y_P=\sqrt{5}-1$, $y_Q=\sqrt{5}+1$.
**Step 3: Compute $d^2$**
$(y_Q-y_P)^2=4$. $x_P^2=\sqrt{5}(\sqrt{5}-1)=5-\sqrt{5}$, $x_Q^2=5+\sqrt{5}$. $(x_Q-x_P)^2=x_Q^2+x_P^2-2\sqrt{x_Q^2x_P^2}=10-2\sqrt{25-5}=10-4\sqrt{5}$. $d^2=(10-4\sqrt{5})+4=14-4\sqrt{5}$.
**Step 4: Extract $a$ and $b$**
$a=14$, $b=-4$. $a-b=14-(-4)=18$.
Correct Answer: 18