JEE Advanced 2026 · Paper 2 · Mathematics Q13

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MathematicsNumerical+4Question 13 of 54

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 $\sqrt5\,y=x^2$ in the first quadrant. Let $d$ be the distance between $P$ and $Q$ .

If $a$ and $b$ are the integers such that $d^2=a+b\sqrt5$ , then the value of $a-b$ is ________.

Your answer

Answer and worked solution

Official answer: 18

Ellipse. $e^2=1-\frac{12}{18}=\frac13$ ; foci $(\pm\sqrt{18}\cdot\frac1{\sqrt3},0)=(\pm\sqrt6,0)$ .

Hyperbola. $e=\sqrt3$ and $ae=\sqrt6$ , so $a^2=2$ , $b^2=a^2(e^2-1)=4$ :
$\frac{x^2}2-\frac{y^2}4=1.$

Intersect with $x^2=\sqrt5\,y$ : $\frac{\sqrt5y}2-\frac{y^2}4=1\implies y^2-2\sqrt5y+4=0\implies y=\sqrt5\pm1$ , both positive, with $x^2=5\pm\sqrt5$ .

Distance. $(y_1-y_2)^2=4$ , and
$(x_1-x_2)^2=x_1^2+x_2^2-2x_1x_2=10-2\sqrt{20}=10-4\sqrt5.$
So $d^2=14-4\sqrt5$ : $a=14$ , $b=-4$ , and $a-b=18$ .

Topics:Hyperbola Ellipse Parabola

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JEE Advanced 2026 · Paper 2, Mathematics question 13