PrepRung
Sign inSign up free

JEE Advanced 2022 Paper 1 · Mathematics · Conic Sections · medium

Match the geometric quantities in List-I with the values in List-II. List-I (P) Length of the chord of the ellipse (x^2/(25)+(y^2/(9)=1 through the…

Single correct option

Match the geometric quantities in List-I with the values in List-II.

List-I
(P) Length of the chord of the ellipse x225+y29=1\dfrac{x^2}{25}+\dfrac{y^2}{9}=1 through the focus (4,0)(4,0) perpendicular to the major axis
(Q) Distance between the foci of the hyperbola x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1
(R) Radius of the circle through (0,0)(0,0), (4,0)(4,0) and (0,3)(0,3)
(S) Area of the triangle bounded by the coordinate axes and the tangent to x2+y2=25x^2+y^2=25 at (3,4)(3,4)

List-II
(1) 52\dfrac52
(2) 1010
(3) 185\dfrac{18}{5}
(4) 62524\dfrac{625}{24}
(5) 77
  1. A. P→3, Q→2, R→1, S→4
  2. B. P→3, Q→5, R→1, S→4
  3. C. P→1, Q→2, R→3, S→4
  4. D. P→3, Q→2, R→4, S→1
Show answer and solution

Answer: (A) P→3, Q→2, R→1, S→4

(P) At x=4x=4: y2=9(1−1625)=8125y^2=9\left(1-\frac{16}{25}\right)=\frac{81}{25}, so the chord is 2⋅95=1852\cdot\frac95=\frac{18}5 (3). (Q) c2=9+16=25c^2=9+16=25, foci at (±5,0)(\pm5,0), distance 1010 (2). (R) The points form a right triangle with hypotenuse from (4,0)(4,0) to (0,3)(0,3) of length 55, which is a diameter, so r=52r=\frac52 (1). (S) The tangent is 3x+4y=253x+4y=25 with intercepts 253\frac{25}3 and 254\frac{25}4, so the area is 12⋅253⋅254=62524\frac12\cdot\frac{25}{3}\cdot\frac{25}{4}=\frac{625}{24} (4). Hence P→3, Q→2, R→1, S→4.
Report a mistake

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake