QJEEAdv23Ma7039: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma7039 Solution: Let and . Three distinct points and are randomly chosen from .…

Question

Verified Solution • 2025–26 Session moderate

Let

X=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}&lt;1\right.

and

\left.y^2&lt;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 :

Accepted Answer

\frac{x^2}{8}+\frac{y^2}{20}&lt;1 \text { and } y^2&lt;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}&amp; \frac{x^2}{8}+\frac{5 x}{20}=1 \\\\&amp; \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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