JEE Advanced 2026 — Paper 2 — MATH-III solution

JEE Advanced2026Paper 2MATH-III

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 $5 y = x^{2}$ in the first quadrant. Let $d$ be the distance between $P$ and $Q$ .

If $a$ and $b$ are the integers such that $d^{2} = a + b 5$ , then the value of $a - b$ is ___________.

Your answer

Solution

Eccentricity of $E$ : $e_{E} = 1 - \frac{12}{18} = \frac{1}{3}$ . Foci of $E$ at $(\pm 18 - 12, 0) = (\pm 6, 0)$ .

Hyperbola $H$ : $e_{H} = 3$ and foci $(\pm 6, 0)$ , so $A \cdot e_{H} = 6$ gives $A = 2$ . Then $B^{2} = A^{2} (e_{H}^{2} - 1) = 2 \cdot 2 = 4$ . Hence $H : \frac{x ^{2}}{2} - \frac{y ^{2}}{4} = 1$ .

Intersection with parabola $x^{2} = 5 y$ : substitute to get $\frac{5 y}{2} - \frac{y ^{2}}{4} = 1$ , i.e. $y^{2} - 25 y + 4 = 0$ .

Let $y_{1}, y_{2}$ be the two $y$ -coordinates. By Vieta: $y_{1} + y_{2} = 25$ , $y_{1} y_{2} = 4$ .

$(y_{1} - y_{2})^{2} = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2} = 20 - 16 = 4$ , so $∣ y_{1} - y_{2} ∣ = 2$ .

Also, $x_{i}^{2} = 5 y_{i}$ , so $x_{1}^{2} + x_{2}^{2} = 5 (y_{1} + y_{2}) = 10$ and $x_{1} x_{2} = x_{1}^{2} x_{2}^{2} = 5 y_{1} y_{2} = 20 = 25$ (positive since both points in the first quadrant).

Therefore $(x_{1} - x_{2})^{2} = x_{1}^{2} + x_{2}^{2} - 2 x_{1} x_{2} = 10 - 45$ .

$d^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = (10 - 45) + 4 = 14 - 45$ .

So $a = 14$ and $b = - 4$ , giving $a - b = 18$ .

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