JEE Main Coordinate Geometry – Topic Test

Topics in this test

Difficulty: 15 medium · 15 hard

Coordinate Geometry MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

1. Q1. Match List I (the conic) with List II (its eccentricity). List I: A. Circle; B. Parabola; C. Ellipse; D. Rectangular hyperbola. List II: 1. e = 1; 2. e = 0; 3. e = √2; 4. 0 < e < 1. Choose the correct matching.

   सूची I (शांकव) का सूची II (इसकी उत्केंद्रता) से मिलान करें। सूची I: A. वृत्त; B. परवलय; C. दीर्घवृत्त; D. समकोणीय अतिपरवलय। सूची II: 1. e = 1; 2. e = 0; 3. e = √2; 4. 0 < e < 1। सही मिलान चुनें।

   • A. A-1, B-2, C-4, D-3
   • B. A-2, B-4, C-1, D-3
   • C. A-2, B-1, C-3, D-4
   • D. A-2, B-1, C-4, D-3
   Solution

   A circle has e=0 (A-2), a parabola e=1 (B-1), an ellipse 0<e<1 (C-4), and a rectangular hyperbola e=√2 (D-3).

   वृत्त की e=0 (A-2), परवलय की e=1 (B-1), दीर्घवृत्त की 0<e<1 (C-4), और समकोणीय अतिपरवलय की e=√2 (D-3) होती है।

2. Q2. The equation of the circle concentric with x² + y² − 4x + 6y − 3 = 0 and passing through the point (5, 1) is:

   वृत्त x² + y² − 4x + 6y − 3 = 0 के संकेंद्रीय (concentric) तथा बिंदु (5, 1) से होकर जाने वाले वृत्त का समीकरण है:

   • A. x² + y² − 4x + 6y − 12 = 0
   • B. x² + y² − 4x + 6y − 3 = 0
   • C. x² + y² + 4x − 6y − 12 = 0
   • D. x² + y² − 4x + 6y + 12 = 0
   Solution

   The common centre is (2, −3); the radius squared is (5−2)² + (1+3)² = 9 + 16 = 25, giving (x−2)² + (y+3)² = 25, i.e. x² + y² − 4x + 6y − 12 = 0.

   उभयनिष्ठ केंद्र (2, −3) है; त्रिज्या का वर्ग (5−2)² + (1+3)² = 9 + 16 = 25 है, जिससे (x−2)² + (y+3)² = 25, अर्थात् x² + y² − 4x + 6y − 12 = 0 प्राप्त होता है।

3. Q3. For the ellipse x²/a² + y²/b² = 1 with a > b, the length of the latus rectum is:

   दीर्घवृत्त x²/a² + y²/b² = 1 (जहाँ a > b) के लिए, नाभिलंब जीवा की लंबाई है:

   • A. a²/b
   • B. b²/2a
   • C. 2b²/a
   • D. 2a²/b
   Solution

   The latus rectum of such an ellipse is 2b²/a.

   ऐसे दीर्घवृत्त की नाभिलंब जीवा 2b²/a होती है।

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