if A= {(x,y)/ x²+y²=25} and B = {(x,y)/ x²+9y²=144} then A intersection B contains
a) one point
b) three points
c) two points
d) four points
Answer is d) four points
Kindly explain
Answers
Given : A = (x,y)(x²+y²=25) and B=(x,y)(x²+9y²=144)
To find : A ∩ B
Solution:
x² + y ² = 25
x² + 9y² = 144
=> 8y² = 119
=> y² = 119/8
=> y = ± √ 119/ 2√2
Substitute y² = 119/8
in x² + y ² = 25
=> x² + 119/8 = 25
=> x² = 81/8
=> x = ± 9/2√2
4 Possible points area
( 9/2√2 , √ 119/ 2√2 ) ,
( 9/2√2 , -√ 119/ 2√2 )
( -9/2√2 , √ 119/ 2√2 )
( -9/2√2 , -√ 119/ 2√2 )
A ∩ B = 4
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Answer:
Clearly, A is the set of all points on the circle x
2
+y
2
=25 and B is the set of all points on the ellipse x
2
+9y
2
=144. These two intersect at four points P,Q,R and S.
Hence, A∩B contains four points.