JEE Advance - Mathematics (2023 - Paper 1 Online - No. 6)
Let $X=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}<1\right.$ and $\left.y^2<5 x\right\}$. Three distinct points $P, Q$ and $R$ are randomly chosen from $X$. Then the probability that $P, Q$ and $R$ form a triangle whose area is a positive integer, is :
$\frac{71}{220}$
$\frac{73}{220}$
$\frac{79}{220}$
$\frac{83}{220}$
Explanation
$$
\frac{x^2}{8}+\frac{y^2}{20}<1 \text { and } y^2<5 x
$$
Let $ \frac{x^2}{8}+\frac{y^2}{20}=1$ ........(1)
and $$ y^2=5 x $$ .........(2)
On solving (1) and (2), we get
$$ \begin{aligned} & \frac{x^2}{8}+\frac{5 x}{20}=1 \\\\ & \frac{x^2}{8}+\frac{x}{4}=1 \end{aligned} $$
$\begin{aligned} x^2+2 x =8 \\\\ \Rightarrow x^2+2 x-8 =0 \\ \\\Rightarrow (x+4)(x-2) =0 \\\\ \Rightarrow x =-4, 2 \\\\ \Rightarrow x =2(-4 \text { is not possible }) \\\\ \Rightarrow y^2 =10 \\\\ \Rightarrow y = \pm \sqrt{10}\end{aligned}$
$\begin{aligned} X=\{(1,1),(1,0),(1,-1),(1,2),(1,-2),(2,1),(2-1), \\ (2,3),(2,3),(2,-3),(2,-2),(2,2),(2,0)\}\end{aligned}$
Let $S$ be the sample space and $E$ be the event $n(S)={ }^{12} C_3$
For E :
Selecting 3 points in which 2 points are either or $\mathrm{x}=1$ and $ \mathrm{x}=2$ but distance between them is even.
Triangles with base 2 :
$$ =3 \times 7+5 \times 5=46 $$
Triangles with base 4 :
$$ =1 \times 7+3 \times 5=22 $$
Triangles with base 6 :
$$ =1 \times 5=5 $$
$P(E)=\frac{46+22+5}{{ }^{12} C_3}=\frac{73}{220}$
Let $ \frac{x^2}{8}+\frac{y^2}{20}=1$ ........(1)
and $$ y^2=5 x $$ .........(2)
On solving (1) and (2), we get
$$ \begin{aligned} & \frac{x^2}{8}+\frac{5 x}{20}=1 \\\\ & \frac{x^2}{8}+\frac{x}{4}=1 \end{aligned} $$
$\begin{aligned} x^2+2 x =8 \\\\ \Rightarrow x^2+2 x-8 =0 \\ \\\Rightarrow (x+4)(x-2) =0 \\\\ \Rightarrow x =-4, 2 \\\\ \Rightarrow x =2(-4 \text { is not possible }) \\\\ \Rightarrow y^2 =10 \\\\ \Rightarrow y = \pm \sqrt{10}\end{aligned}$
$\begin{aligned} X=\{(1,1),(1,0),(1,-1),(1,2),(1,-2),(2,1),(2-1), \\ (2,3),(2,3),(2,-3),(2,-2),(2,2),(2,0)\}\end{aligned}$
Let $S$ be the sample space and $E$ be the event $n(S)={ }^{12} C_3$
For E :
Selecting 3 points in which 2 points are either or $\mathrm{x}=1$ and $ \mathrm{x}=2$ but distance between them is even.
Triangles with base 2 :
$$ =3 \times 7+5 \times 5=46 $$
Triangles with base 4 :
$$ =1 \times 7+3 \times 5=22 $$
Triangles with base 6 :
$$ =1 \times 5=5 $$
$P(E)=\frac{46+22+5}{{ }^{12} C_3}=\frac{73}{220}$
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