JEE Main 2022 — Ellipse Question with Solution

Given,

$S: 9(x-3{)}^2+16{\left(y-4\right)}^2\le 144 \text{where }\left(x,y\right)\in N\times N$

On rearranging we get,

$S:\frac{(x-3{)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1 \text{where }x,y\in \left\{1,2,3,\dots \dots \right\}$

Now $T:{\left(x-7\right)}^2+{\left(y-4\right)}^2\le 36 \text{where }x,y\in R$

Now let $x-3=X:y-4=Y$

So, new equation and domain will become,

$S:\frac{X^2}{16}+\frac{Y^2}{9}\le 1; X\in \left\{-2,-1,0,1,\dots \dots \right\}$

$T:{\left(X-4\right)}^2+Y^2\le 36; \mathrm{Y}\in \left\{-3,-2,-1,0,\dots \dots \right\}$

Now on plotting the diagram we get,

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

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

So, possible for $S\cap T$ will be,

When $Y=0$$\left(-2,0\right),\left(-1,0\right),\dots \dots \left(4,0\right)\to 7 \text{possible case}$

When $Y=1$ 

$\left(-1,1\right),\left(0,1\right),\dots \dots .\left(3,1\right)\to 5 \text{possible case}$

When $Y=-1$ 

$\left(-1,-1\right),\left(0,-1\right),\dots \dots .\left(3,-1\right)\to 5 \text{possible case}$

When $Y=2$ 

$\left(-1,2\right),\left(0,2\right),\left(1,2\right),\left(2,2\right)\to 4 \text{possible case}$

When $Y=-2$

$\left(-1,-2\right),\left(0,-2\right),\left(1,-2\right),\left(2,-2\right)\to 4 \text{possible case}$

When $Y=3 \& -3$

$\left(0,3\right)\left(0,-3\right)\to \text{Total }2 \text{possible case}$

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