Question

if A = { x^2 + y^2 = 16 } and B = { 9x^2 + 25y^2 = 225 } Find n(A intersection B) ​

Answer 1

Answer:

$\boxed{\red{\sf n( A \cap B)= \boxed{4}}}$

Step-by-step explanation:

Two sets are given to us and we need to find the number of elements in the intersection of the two sets . The given sets are ,

$\sf\red{\dashrightarrow} A = \{ x^2 + y^2= 16 \}$

$\sf\red{\dashrightarrow} B = \{ 9x^2 + 25y^2= 225 \}$

• The equation in the first set is ,

$\sf\dashrightarrow x^2 + y^2= 16 \\\\\sf\dashrightarrow\boxed{\red{ \sf( x - 0)^2 + ( y-0)^2 = 4^2 }}$

• This is similar to the standard equation of circle. The Standard equⁿ of circle is ( x - h)² + ( y - k)² = r² .
• Where ( h , k) is the centre and r is the radius of the circle.

Therefore the centre of the circle is (0,0) with radius 4 units . Hence the circle will cut the x - axis at (-4,0) and (4,0) .

• The equation of second set is ,

$\sf\dashrightarrow 9x^2 + 25y^2 = 225 \\\\\sf\dashrightarrow \dfrac{9x^2+25y^2}{225}= 1 \\\\\sf\dashrightarrow \dfrac{9x^2}{225}+\dfrac{25y^2}{225}=1 \\\\\sf\dashrightarrow \dfrac{x^2}{25}+\dfrac{y^2}{9}= 1 \\\\\sf\dashrightarrow \boxed{\red{\sf \dfrac{(x-0)^2}{25}+\dfrac{(y-0)^2}{9}= 1 }}$

• This is similar to the standard equation of eclipse . The Standard equation of Eclipse is (x-h)²/a² + (y-k)²/b² = 1 .
• Where (h,k) is the centre .

The equation will cut x axis on (5,0) and (-5,0) also it will cut y axis on (-3,0) and (3,0) .

• The point on the graph where both the circle and the eclipse will cut each other will give us the number of elements in intersection of sets .
• From the graph , both of them intersect each other at four distinct points . Therefore the number of elements in the intersection of sets will be four .

• Also Refer to both attachments .

$\rule{200}5$

$\qquad\qquad\sf\dashrightarrow \boxed{\boxed{\pink{\sf n(A\cap B)= 4 }}}$

$\rule{200}5$

Answer 2

Answer:

Answer:

\boxed{\red{\sf n( A \cap B)= \boxed{4}}}

n(A∩B)=

4

Step-by-step explanation:

Two sets are given to us and we need to find the number of elements in the intersection of the two sets . The given sets are ,

\sf\red{\dashrightarrow} A = \{ x^2 + y^2= 16 \} ⇢A={x

2

+y

2

=16}

\sf\red{\dashrightarrow} B = \{ 9x^2 + 25y^2= 225 \} ⇢B={9x

2

+25y

2

=225}

• The equation in the first set is ,

\begin{gathered}\sf\dashrightarrow x^2 + y^2= 16 \\\\\sf\dashrightarrow\boxed{\red{ \sf( x - 0)^2 + ( y-0)^2 = 4^2 }}\end{gathered}

⇢x

2

+y

2

=16

⇢

(x−0)

2

+(y−0)

2

=4

2

This is similar to the standard equation of circle. The Standard equⁿ of circle is ( x - h)² + ( y - k)² = r² .

Where ( h , k) is the centre and r is the radius of the circle.

Therefore the centre of the circle is (0,0) with radius 4 units . Hence the circle will cut the x - axis at (-4,0) and (4,0) .

• The equation of second set is ,

\begin{gathered}\sf\dashrightarrow 9x^2 + 25y^2 = 225 \\\\\sf\dashrightarrow \dfrac{9x^2+25y^2}{225}= 1 \\\\\sf\dashrightarrow \dfrac{9x^2}{225}+\dfrac{25y^2}{225}=1 \\\\\sf\dashrightarrow \dfrac{x^2}{25}+\dfrac{y^2}{9}= 1 \\\\\sf\dashrightarrow \boxed{\red{\sf \dfrac{(x-0)^2}{25}+\dfrac{(y-0)^2}{9}= 1 }}\end{gathered}

⇢9x

2

+25y

2

=225

⇢

225

9x

2

+25y

2

=1

⇢

225

9x

2

+

225

25y

2

=1

⇢

25

x

2

+

9

y

2

=1

⇢

25

(x−0)

2

+

9

(y−0)

2

=1

This is similar to the standard equation of eclipse . The Standard equation of Eclipse is (x-h)²/a² + (y-k)²/b² = 1 .

Where (h,k) is the centre .

The equation will cut x axis on (5,0) and (-5,0) also it will cut y axis on (-3,0) and (3,0) .

The point on the graph where both the circle and the eclipse will cut each other will give us the number of elements in intersection of sets .

From the graph , both of them intersect each other at four distinct points . Therefore the number of elements in the intersection of sets will be four .

Also Refer to both attachments .

\rule{200}5

\qquad\qquad\sf\dashrightarrow \boxed{\boxed{\pink{\sf n(A\cap B)= 4 }}}⇢

n(A∩B)=4

\rule{200}5

Step-by-step explanation:

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