Let , and . Let A(4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and X(3r – n, – n – 1) be the vertices of a triangle ABC, where t is a parameter. If , is the locus of the centroid of triangle ABC, then equals [2025]

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Solution

Q.
Let ${}^{n}C_{r - 1} = 28$ , ${}^{n}C_{r} = 56$ and ${}^{n}C_{r + 1} = 70$ . Let A(4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and C(3r – n, $r^{2}$ – n – 1) be the vertices of a triangle ABC, where t is a parameter. If ${(3 x - 1)}^{2} + {(3 y)}^{2} = α$ , is the locus of the centroid of triangle ABC, then $α$ equals [2025]

1 18

2 8

3 20

4 6

Ans.
(3)

We have, ${}^{n}C_{r - 1} = 28$ , ${}^{n}C_{r} = 56$ and ${}^{n}C_{r + 1} = 70$

$\frac{{}^{n}C_{r - 1}}{{}^{n}C_{r}} = \frac{28}{56} \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} \times \frac{r! (n - r)!}{n!} = \frac{1}{2}$

$\Rightarrow \frac{r}{(n - r + 1)} = \frac{1}{2} \Rightarrow 3 r = n + 1$ ... (i)

Also, $\frac{{}^{n}C_{r}}{{}^{n}C_{r + 1}} = \frac{56}{70} \Rightarrow \frac{r + 1}{n - r} = \frac{4}{5} \Rightarrow 9 r = 4 n - 5$ ... (ii)

From equations (i) and (ii), we get r = 3 and n = 8.

Now, A = (4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and C(1, 0)

$∴$ Centroid of triangle is

$(\frac{4 c o s t + 2 \sin t + 1}{3}, \frac{4 \sin t - 2 \cos t}{3})$

Then, ${(3 x - 1)}^{2} + {(3 y)}^{2} = {(4 \cos t + 2 \sin t)}^{2} + {(4 \sin t - 2 \cos t)}^{2} {(3 x - 1)}^{2} + {(3 y)}^{2} = 20$

So, value of $α = 20$ .

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