Let ^ n C_ r-1 =28, ^ n C_ r =56 and ^ n C_ r+1 =70 . Let A(4 t, 4 t), B(2 t,-2 t) and C(3 r-n, r^2-n-1) be the vertices of a triangle A B C , where t is a parameter. If (3 x-1)^2+(3 y)^2 =α , is the locus of the

Correct answer: D. 20 . array l ^n C_ r-1 =28 ^n C_r=56 ^n C_ r+1 =70 array l ^n C_ r-1 ^n C_r array =(28)/(56) (r)/(n-r+1)=(1)/(2) ^n C_r ^n C_ r+1 =(56)/(70) (r+1)/(n-r)=(70)/(56) array \ n=8=3 h=(4 t+2 t+1)/(3) k=(4 t-2 t)/(3) 3 h-1=4 t+2 t 3 k-1=4 t-2 t aligned & (1)^2+(2)^2 & (3 h-1)^2+(3 k)^2=20 aligned Locus of centroid:(3 x-1)^2+(3 y)^2=20 α=20

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Straight Lines

Let ^ n C_ r-1 =28, ^ n C_ r =56 and ^ n C_ r+1 =70 . Let A(4 t, 4...

Let

{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56 and

{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}=70

. Let

\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos \mathrm{t})

and

C\left(3 r-n, r^2-n-1\right)

be the vertices of a triangle

A B C

, where

t

is a parameter. If

(3 x-1)^2+(3 y)^2

=\alpha

, is the locus of the centroid of triangle ABC , then

\alpha

equals

Let ^ n C_ r-1 =28, ^ n C_ r =56 and ^ n C_ r+1 =70 . Let A(4 t, 4 t), B(2 t,-2 t) and C(3 r-n, r^2-n-1) be the

Options

Correct answer

D. 20

Step-by-step solution

\left.

\begin{array}{l}{ }^n C_{r-1}=28 \\ { }^n C_r=56 \\ { }^n C_{r+1}=70 \\ \begin{array}{l}{ }^n C_{r-1} \\ { }^n C_r\end{array}

=\frac{28}{56} \Rightarrow \frac{r}{n-r+1}=\frac{1}{2} \\ \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \Rightarrow \frac{r+1}{n-r}=\frac{70}{56}\end{array}

\right\} n=8=3

h=\frac{4 \cos t+2 \sin t+1}{3}

\quad

k=\frac{4 \sin t-2 \cos t}{3} 3

h-1=4 \cos t+2 \sin t 3 k-1=4 \sin t-2 \cos t

\begin{aligned} & (1)^2+(2)^2 \\ & (3 h-1)^2+(3 k)^2=20 \end{aligned}

$Locus of centroid:(3 x-1)^2+(3 y)^2=20 \Rightarrow \alpha=20

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