Straight Lines Question - None

Question

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 ${C}\left(3 r-n, r^{2}-n-1\right)$ be the vertices of a triangle ABC, 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

Given,
${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56,{ }^{n} C_{r+1}=70$
$\boldsymbol{A}({4} \cos {t}, {4} \sin t)$,
$B(2 \sin t,-2 \cos t)$ and $C\left(3 r-n, r^{2}-n-1 ight)$ are vertices of $ riangle A B C$.
Now, $\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{28}{56}$
$\frac{r}{(n-r+1)}=\frac{1}{2}$
$\Rightarrow$ ${3 r}=\boldsymbol{n}+{1}$ ...(i)
Similarly, $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{56}{70}$
$\Rightarrow \frac{(r+1)}{(n-r)}=\frac{56}{70}$
$\Rightarrow 9 r=4 n-5$ ...(ii)
On simplifying (i) & (ii), we get
$r=3, n=8$
Now the vertices are
$A(4 \cos t, 4 \sin t), B(2 \sin t,-2 \cos t), C(1,0)$

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

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