If A = [(x,,y):{x^2} + {y^2} = 25] and B = [( | vedclass

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(d) $A = $ Set of all values $(x, y) :$ ${x^2} + {y^2} = 25 = {5^2}$

$B = {{{x^2}} \over {144}} + {{{y^2}} \over {16}} = 1$ i.e., ${{{x^2}} \over {

Vedclass · Accepted Answer

Four points

Vedclass · Answer

(d) $A = $ Set of all values $(x, y) :$ ${x^2} + {y^2} = 25 = {5^2}$

$B = {{{x^2}} \over {144}} + {{{y^2}} \over {16}} = 1$ i.e., ${{{x^2}} \over {{{(12)}^2}}}$ + ${{{y^2}} \over {{{(4)}^2}}} = 1$.

Clearly, $ A \cap 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

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