If A - x-y- x2 - y2 - 25 And B - x-y- x2 - 9y2 - 144- then A - B contains

A = Set of all values (x,y) : x^2 + y^2 = 25 = 5^2 nbsp; and nbsp; B = Set of all values (x,y): nbsp;x^2/144 + y^2/16 = 1 nbsp; nbsp;i.e., x^2/(12) ...

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Byju's Answer

Standard XIII

Mathematics

Intersection

If A = x,y: x...

Question

If A = { $(x, y)$ : $x^{2} + y^{2} = 25$ } And B = {(x,y): $x^{2} + 9 y^{2} = 144$ }, then $A \cap B$ contains

One point

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Three points

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Two points

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Four points

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Solution

The correct option is D
Four points

A = Set of all values (x,y) : $x^{2} + y^{2} = 25 = 5^{2}$
and

B = Set of all values (x,y): $\frac{x^{2}}{144} + \frac{y^{2}}{16} = 1$ i.e., $\frac{x^{2}}{(12)^{2}} + \frac{y^{2}}{(4)^{2}} = 1$

Clearly , A $\cap$ B consists of four points.

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Similar questions

If A = [(x,y): $x^{2} + y^{2} = 25$ ] And B = [(x,y): $x^{2} + 9 y^{2} = 144], t h e n A \cap B$ contains

Q. If

A = {(x, y) : x^{2} + y^{2} = 25}

and

B = {(x, y) : x^{2} + 9 y^{2} = 144}

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A \cap B

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If $A = [(x, y) : x^{2} + y^{2} = 25]$ And $B = [(x, y) : x^{2} + 9 y^{2} = 144], t h e n A \cap B$ contains

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and

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x

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are both positive?

Q. If

\frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y}

then prove that.

\frac{a (b - c)}{y^{2} - z^{2}} = \frac{b (c - a)}{z^{2} - x^{2}} = \frac{c (a - b)}{x^{2} - y^{2}}

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If A = { (x,y): x2+y2=25 } And B = {(x,y): x2+9y2=144}, then A∩B contains

The correct option is D Four points A = Set of all values (x,y) : x2+y2=25=52 and B = Set of all values (x,y): x2144+y216=1 i.e., x2(12)2+y2(4)2=1 Clearly , A ∩ B consists of four points.

If A = { $(x, y)$ : $x^{2} + y^{2} = 25$ } And B = {(x,y): $x^{2} + 9 y^{2} = 144$ }, then $A \cap B$ contains

The correct option is D
Four points

A = Set of all values (x,y) : $x^{2} + y^{2} = 25 = 5^{2}$
and

B = Set of all values (x,y): $\frac{x^{2}}{144} + \frac{y^{2}}{16} = 1$ i.e., $\frac{x^{2}}{(12)^{2}} + \frac{y^{2}}{(4)^{2}} = 1$

Clearly , A $\cap$ B consists of four points.