JEE Main 2025 — 28 January, Morning Shift — Question 17

${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$

$\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56}$

$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2}$

$\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2}$

$3 \mathrm{r}=\mathrm{n}+1$

$\frac{(\mathrm{r}+1)}{(\mathrm{n}-\mathrm{r})}=\frac{56}{70} \Rightarrow 9 \mathrm{r}=4 \mathrm{n}-5$

By (i) & (ii)

$(\mathrm{r}=3),(\mathrm{n}=8)$

A (4cost, 4sint), B(2sint, $-2 \cos t), C\left(3 r-n, r^{2}-n-1\right)$

$\mathrm{A}(4 \cos t, 4 \sin t) \quad \mathrm{B}(2 \sin t,-2 \cos t) \quad \mathrm{C}(1,0)$

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

$(3 x-1)^{2}+(3 y)^{2}=20$