JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 21)
$$\text { Let } S=\left\{(x, y) \in \mathbb{N} \times \mathbb{N}: 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$$ and
$$T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$$. Then $$n(S \cap T)$$ is equal to __________.
Answer
27
Explanation
$S=\left\{(x, y) \in \mathbb{N} \times \mathbb{N}: \frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9} \leq 1\right\}$
represents all the integral points inside
and on the ellipse $\frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9}=1$, in first quadrant.
and $T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ represents all the points on
and inside the circle $(x-7)^{2}+(y-4)^{2}=36$
_29th_July_Evening_Shift_en_21_1.png)
$\therefore \quad n(S \cap T)=\{(3,1),(2,2),(3,2),(4,2),(5,2)$, $(2,3), \ldots(6,5)\}$
Total number of points $=27$
represents all the integral points inside
and on the ellipse $\frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9}=1$, in first quadrant.
and $T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ represents all the points on
and inside the circle $(x-7)^{2}+(y-4)^{2}=36$
$\therefore \quad n(S \cap T)=\{(3,1),(2,2),(3,2),(4,2),(5,2)$, $(2,3), \ldots(6,5)\}$
Total number of points $=27$
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