JEE Main 2022 — Mathematics Coordinate Geometry

Given,

$$S:9(x-3{)}^{2}+16{(y-4)}^{2}\leq 144\text{where }(x,y)\in N\times N$$S:9(x−3)2+16(y−4)2≤144where (x,y)∈N×N

On rearranging we get,

$$S:\frac{(x-3{)}^{2}}{16}+\frac{{(y-4)}^{2}}{9}\leq 1\text{where }x,y\in {1,2,3,\ldots \ldots }$$S:16(x−3)2​+9(y−4)2​≤1where x,y∈1,2,3,……

Now $T:{(x-7)}^{2}+{(y-4)}^{2}\leq 36\text{where }x,y\in R$T:(x−7)2+(y−4)2≤36where x,y∈R

Now let $x-3=X:y-4=Y$x−3=X:y−4=Y

So, new equation and domain will become,

$$S:\frac{{X}^{2}}{16}+\frac{{Y}^{2}}{9}\leq 1;X\in {-2,-1,0,1,\ldots \ldots }$$S:16X2​+9Y2​≤1;X∈−2,−1,0,1,……

$$T:{(X-4)}^{2}+{Y}^{2}\leq 36;Y\in {-3,-2,-1,0,\ldots \ldots }$$T:(X−4)2+Y2≤36;Y∈−3,−2,−1,0,……

Now on plotting the diagram we get,

Now number of point in $n(S\cap T)$n(S∩T) will be given by,

Take $Y=-3,-2,-1,0,1,2,3$Y=−3,−2,−1,0,1,2,3 and check for $X$X

So, possible for $S\cap T$S∩T will be,

When $Y=0$Y=0$(-2,0),(-1,0),\ldots \ldots (4,0)\rightarrow 7\text{possible case}$(−2,0),(−1,0),……(4,0)→7possible case

When $Y=1$Y=1

$$(-1,1),(0,1),\ldots \ldots .(3,1)\rightarrow 5\text{possible case}$$(−1,1),(0,1),…….(3,1)→5possible case

When $Y=-1$Y=−1

$$(-1,-1),(0,-1),\ldots \ldots .(3,-1)\rightarrow 5\text{possible case}$$(−1,−1),(0,−1),…….(3,−1)→5possible case

When $Y=2$Y=2

$$(-1,2),(0,2),(1,2),(2,2)\rightarrow 4\text{possible case}$$(−1,2),(0,2),(1,2),(2,2)→4possible case

When $Y=-2$Y=−2

$$(-1,-2),(0,-2),(1,-2),(2,-2)\rightarrow 4\text{possible case}$$(−1,−2),(0,−2),(1,−2),(2,−2)→4possible case

When Y=3&-3

$$(0,3)(0,-3)\rightarrow \text{Total }2\text{possible case}$$(0,3)(0,−3)→Total 2possible case

So, total cases will be $7+5+5+4+4+2=27$7+5+5+4+4+2=27