Sets & RelationsmediumFree

Intersection of Circle x²+y²=25 and Ellipse | JEE Sets

JEE Maths question with a full step-by-step solution.

Question
If A={(x,y):x2+y2=25}A = \{(x, y) : x^2 + y^2 = 25\} and B={(x,y):x2+9y2=144}B = \{(x, y) : x^2 + 9y^2 = 144\}, then A∩BA \cap B contains:
AOne point
BThree points
CTwo points
DFour pointscorrect
Solution
AA is the circle x2+y2=25x^2 + y^2 = 25 and BB is the ellipse x2144+y216=1\dfrac{x^2}{144} + \dfrac{y^2}{16} = 1. Step 1: From BB, x2=144−9y2x^2 = 144 - 9y^2. Substitute into AA:
144−9y2+y2=25⇒144−8y2=25⇒y2=1198.144 - 9y^2 + y^2 = 25 \Rightarrow 144 - 8y^2 = 25 \Rightarrow y^2 = \frac{119}{8}.
 Step 2:
x2=25−y2=25−1198=818>0.x^2 = 25 - y^2 = 25 - \frac{119}{8} = \frac{81}{8} > 0.
 Two values of yy and two of x⇒4x \Rightarrow 4 intersection points. Correct answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sets & Relations · easy
The relation R = (1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3) on the set A = 1, 2, 3 is:
Sets & Relations · easy
The relation R is defined in A = 1, 2, 3 by aRb if |a² - b²| ≤ 5 . Which of the following…
Sets & Relations · medium
The relation R defined on the set A = 1, 2, 3, 4, 5 by R = (x, y) : |x² - y²| < 16 is giv…
Sets & Relations · easy
If A is the set of even natural numbers less than 8 and B is the set of prime numbers les…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath