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)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\} on the set A={1,2,3}A = \{1, 2, 3\} is:
Sets & Relations · easy
The relation RR is defined in A={1,2,3}A = \{1, 2, 3\} by aRbaRb if ∣a2−b2∣≤5|a^2 - b^2| \le 5. Which of the following is false?
Sets & Relations · medium
The relation RR defined on the set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} by R={(x,y):∣x2−y2∣<16}R = \{(x, y) : |x^2 - y^2| < 16\} is given by:
Sets & Relations · easy
If AA is the set of even natural numbers less than 88 and BB is the set of prime numbers less than 77, then the number of relations from AA to BB is:

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