JEE Advanced 2026 · Paper 1 · Mathematics Q16

← JEE Advanced 2026 · Paper 1

MathematicsList match+4 / -1Question 16 of 48

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

List-I

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

List-II

1
the point $(1,1)$
2
the point $(7,9)$
3
the point $(3,2)$
4
the point $(2,5)$
5
the point $\left(8,3\sqrt3\right)$

P → 3; Q → 4; R → 1; S → 2

P → 3; Q → 2; R → 1; S → 5

P → 3; Q → 2; R → 4; S → 5

P → 4; Q → 1; R → 2; S → 3

Answer and worked solution

Official answer: B

(P) Radius $=\dfrac{|3+8-1|}{5}=2$ , so the circle is $(x-1)^2+(y-2)^2=4$ . The point $(3,2)$ gives $4+0=4$ → 3.

(Q) A tangent to $y^2=8x$ is $y=mx+\frac2m$ . Tangency to $x^2+y^2=2$ : $\frac{4/m^2}{1+m^2}=2\implies m^4+m^2-2=0\implies m=1$ . The line $y=x+2$ passes through $(7,9)$ → 2.

(R) $\frac{x^2}{16}+\frac{y^2}{12}=1$ : $e=\frac12$ , focus $(2,0)$ , semi-latus rectum $\frac{12}4=3$ , so $M=(2,3)$ . Normal: $\frac{16x}{2}-\frac{12y}{3}=16-12$ , i.e. $2x-y=1$ , which passes through $(1,1)$ → 1.

(S) $ae=5$ and $\frac ae=\frac{16}5$ give $a=4$ , $e=\frac54$ , $b^2=16\left(\frac{25}{16}-1\right)=9$ . Check $\left(8,3\sqrt3\right)$ : $\frac{64}{16}-\frac{27}{9}=1$ → 5.

P→3, Q→2, R→1, S→5.

Topics:Tangents and normals to conics Equation of a circle Ellipse Hyperbola

← Previous question Next question →

JEE Advanced 2026 · Paper 1, Mathematics question 16