Let ^n C_r-1=28, ^n C_r=56 and ^n C_r+1=70.

Question

Let ${ }^n C_{r-1}=28,{ }^n C_r=56$ and ${ }^n C_{r+1}=70$. Let $A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)$ 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

Accepted Answer

$$\begin{aligned}
& { }^n C_{r-1}=28,{ }^n C_r=56 \
& \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56} \
& \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \
& \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2} \
& 3 \mathrm{r}=\mathrm{n}+1\quad\text{..... (i)}
\end{aligned}$$

$$\begin{aligned}
& \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \
& \frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5\quad\text{.... (ii)}
\end{aligned}$$

$$\begin{aligned}
& \text { By (i) & (ii) } \
& \[\begin{gathered} (r=3),(n=8) \ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right) \ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \operatorname{cost}) C(1,0) \ (3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \sin t)^2+(4 \sin t-\operatorname{cost})^2 \ (3 x-1)^2+(3 y)^2=20 \quad \therefore \text { option }(1) \end{gathered}\end{aligned}$$
\]

Solutions · 1