JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Hyperbola

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 ____.

Answer: 18

Solution

Ellipse E:

e_{E} = 1 - \frac{12}{18} = \frac{1}{3}

, foci at

(\pm 6, 0)

. Hyperbola H has

e_{H} = 3

, same foci. Thus

A^{2} + B^{2} = 6

and

A e_{H} = 6

, giving

A = 2, B^{2} = 4

. So H:

\frac{x ^{2}}{2} - \frac{y ^{2}}{4} = 1

. Substituting

x^{2} = 5 y

into H:

\frac{5 y}{2} - \frac{y ^{2}}{4} = 1 \Rightarrow y^{2} - 25 y + 4 = 0

. Roots

y_{1}, y_{2}

satisfy

y_{1} + y_{2} = 25

y_{1} y_{2} = 4

. Distance

d^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}

. Since

(y_{1} - y_{2})^{2} = 20 - 16 = 4

and

x_{1}^{2} + x_{2}^{2} - 2 x_{1} x_{2} = 5 (y_{1} + y_{2}) - 2 5 \cdot 4 = 10 - 45

, we get

d^{2} = 10 - 45 + 4 = 14 - 45

. Thus

a = 14, b = - 4

, so

a - b = 18

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

Practise more: all JEE Advanced PYQs on Hyperbola →

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Hyperbola

Solution

More JEE Advanced questions on Hyperbola