JEE Advanced 2026 — Paper 1 — MATH-IV solution

JEE Advanced2026Paper 1MATH-IV

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I	List-II
(P) The circle with centre $(1, 2)$ and touching the straight line $3 x + 4 y = 1$ , passes through	(1) the point $(1, 1)$
(Q) The common tangent to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ with positive slope, passes through	(2) the point $(7, 9)$
(R) Let M be the end point of the latus rectum of the ellipse $3 x^{2} + 4 y^{2} = 48$ such that M lies in the first quadrant. Then the normal to the ellipse drawn at M passes through	(3) the point $(3, 2)$
(S) Let H be the hyperbola whose centre is at the origin, one of the foci is at $(5, 0)$ , and one directrix is $5 x + 16 = 0$ . Then H passes through	(4) the point $(2, 5)$
	(5) the point $(8, 33)$

A
(P) $\to$ (3), (Q) $\to$ (4), (R) $\to$ (1), (S) $\to$ (2)
B
(P) $\to$ (3), (Q) $\to$ (2), (R) $\to$ (1), (S) $\to$ (5)
C
(P) $\to$ (3), (Q) $\to$ (2), (R) $\to$ (4), (S) $\to$ (5)
D
(P) $\to$ (4), (Q) $\to$ (1), (R) $\to$ (2), (S) $\to$ (3)

Solution

(P) Radius $=$ distance from centre $(1, 2)$ to the line $3 x + 4 y - 1 = 0$ :

$r = \frac{∣3 + 8 - 1∣}{9 + 16} = \frac{10}{5} = 2$

Circle: $(x - 1)^{2} + (y - 2)^{2} = 4$ . Checking the listed points, $(3, 2)$ gives $4 + 0 = 4$ . P $\to$ 3.

(Q) For $y^{2} = 8 x$ (with $4 a = 8$ , $a = 2$ ), a tangent of slope $m$ is $y = m x + \frac{2}{m}$ . For this line to also touch $x^{2} + y^{2} = 2$ , the perpendicular distance from origin equals $2$ :

$\frac{∣2/ m ∣}{1 + m ^{2}} = 2 \Rightarrow \frac{4}{m ^{2}} = 2 (1 + m^{2}) \Rightarrow m^{4} + m^{2} - 2 = 0 \Rightarrow m^{2} = 1$

Taking $m = 1$ (positive slope): $y = x + 2$ . Checking, $(7, 9)$ satisfies $9 = 7 + 2$ . Q $\to$ 2.

(R) Ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1$ gives $a^{2} = 16$ , $b^{2} = 12$ , $e^{2} = 1 - 12/16 = 1/4$ , so $e = 1/2$ and $a e = 2$ . End of latus rectum in the first quadrant: $M = (a e, \frac{b ^{2}}{a}) = (2, 3)$ .

Normal at $(x_{1}, y_{1})$ on $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is $\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}$ . Substituting: $\frac{16 x}{2} - \frac{12 y}{3} = 4$ , i.e. $8 x - 4 y = 4$ , or $2 x - y = 1$ . This passes through $(1, 1)$ . R $\to$ 1.

(S) Focus $(5, 0)$ gives $a e = 5$ . Directrix $x = - 16/5$ gives $a / e = 16/5$ . Multiplying: $a^{2} = 16$ , so $a = 4$ and $e = 5/4$ . Then $b^{2} = a^{2} (e^{2} - 1) = 16 (25/16 - 1) = 9$ .

Hyperbola: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ . Testing $(8, 33)$ : $\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1$ . $✓$ S $\to$ 5.

$∴$ The correct option is (B).

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