Let ⁿC_{r-1}=28, ⁿC_r=56 and ⁿC_{r+1}=70 Let A(4 cos t, 4 sin t), B(2 sin t, -2 cos t) and C(3r-n, r

Q: Wait, why do we reject the negative root here?

Actually, in this problem, we don't have a quadratic in $n$ or $r$ that forces a rejection. We have a system of linear equations. If you ever encounter a quadratic, remember that $n$ must be a positive integer and $r$ must satisfy $0 \le r \le n$. Always check these constraints!

Q: Why can't we just use energy conservation?

Energy conservation is for physics! Here, we are dealing with coordinate geometry and locus. The 'energy' of this problem lies in the invariance of the distance from the origin as the parameter $t$ varies. We are essentially tracing a circle.

Q: Is there a faster way to find the locus without squaring?

You could use rotation matrices or complex numbers, but squaring and adding is the standard, most robust method for linear combinations of $\sin t$ and $\cos t$. It is the 'golden key' for these types of JEE problems.

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 $C(3r-n,r^{2}-n-1)$ be the vertices of a triangle ABC, where t is a parameter. If $(3x-1)^{2}+(3y)^{2}=\alpha.$ is the locus of the centroid of triangle ABC, then $\alpha$ equals

Options:

(A) 6

(B) 18

(C) 8

(D) 20

Sigma Percentile · Accepted Answer

### Analyzing the Binomial Coefficients

To solve this problem, we first determine the values of $n$ and $r$ using the properties of consecutive binomial coefficients. We are given:
$${}^{n}C_{r-1} = 28, \quad {}^{n}C_{r} = 56, \quad {}^{n}C_{r+1} = 70$$

We utilize the ratio property of binomial coefficients, which states that $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$. Substituting the given values:
$$\frac{56}{28} = 2 \Rightarrow \frac{n-r+1}{r} = 2 \Rightarrow n-r+1 = 2r \Rightarrow n = 3r-1$$

Next, we apply the ratio property to the second pair: $\frac{{}^{n}C_{r+1}}{{}^{n}C_{r}} = \frac{n-r}{r+1}$. Substituting the values:
$$\frac{70}{56} = \frac{5}{4} \Rightarrow \frac{n-r}{r+1} = \frac{5}{4} \Rightarrow 4n-4r = 5r+5 \Rightarrow 4n = 9r+5$$

By substituting $n = 3r-1$ into the second equation, we get $4(3r-1) = 9r+5$, which simplifies to $12r-4 = 9r+5$. Solving this yields **$r=3$** and **$n=8$**.

### The Geometry of the Triangle

With $n=8$ and $r=3$, we determine the coordinates of vertex $C$. Given $C(3r-n, r^2-n-1)$, we substitute the values:
$$C(3(3)-8, 3^2-8-1) = C(9-8, 9-8-1) = C(1, 0)$$

The vertices of the triangle are $A(4\cos t, 4\sin t)$, $B(2\sin t, -2\cos t)$, and $C(1, 0)$. The centroid $G(x, y)$ is the average of the vertices:
$$x = \frac{4\cos t + 2\sin t + 1}{3}, \quad y = \frac{4\sin t - 2\cos t + 0}{3}$$

### The Locus of the Centroid

To find the locus, we isolate the trigonometric terms:
$$3x-1 = 4\cos t + 2\sin t$$
$$3y = 4\sin t - 2\cos t$$

We eliminate the parameter $t$ by squaring and adding both equations:
$$(3x-1)^2 = (4\cos t + 2\sin t)^2 = 16\cos^2 t + 4\sin^2 t + 16\sin t \cos t$$
$$(3y)^2 = (4\sin t - 2\cos t)^2 = 16\sin^2 t + 4\cos^2 t - 16\sin t \cos t$$

Adding these expressions causes the cross-terms to cancel:
$$(3x-1)^2 + (3y)^2 = 16(\cos^2 t + \sin^2 t) + 4(\sin^2 t + \cos^2 t)$$
$$(3x-1)^2 + (3y)^2 = 16(1) + 4(1) = 20$$

The locus of the centroid is $(3x-1)^2 + (3y)^2 = 20$. Thus, the final value is **$\alpha = 20$**.

Select Answer:

Visualized Solution

Ratio of Binomial Coefficients

First Linear Equation

Second Ratio

Second Linear Equation

Finding $n$ and $r$

Coordinates of Vertex $C$

Centroid of $Δ A B C$

Expressing $x$ and $y$

Isolating Trigonometric Terms

Eliminating Parameter $t$

Expanding Squares

Finding $α$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Binomial Coefficients

The Geometry of the Triangle

The Locus of the Centroid

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Select Answer:

Visualized Solution

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