Q82 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics
For a real variable $a>1$, consider the points $A_k=\left(k a, a^k\right), k=1,2, \ldots ., n$ in the Cartesian plane. If $\alpha$ and $\beta$ represent respectively the arithmetic mean of $x$-coordinates and the geometric mean of $y$ coordinates of $A_k$, then the locus of the point $P(\alpha, \beta)$ is
- A$n y=\left(\frac{2 x}{n}\right)^{n^2+1}$
- Correct answerB$y^2=\left(\frac{2 x}{n+1}\right)^{n+1}$
- C$y=\left(\frac{x^2}{n+1}\right)^n$
- D$y=(n+1)(x-(n+1))$
Solution
We have,
$\begin{aligned} \alpha & =\frac{a+2 a+3 a+\ldots+n a}{n} \\ & =\frac{a[1+2+3+\ldots+n]}{n} \\ & =\frac{a n(n+1)}{2 n}=\frac{a(n+1)}{2}\end{aligned}$
$\begin{aligned} & \text { and } \beta=\left(a \cdot a^2 \cdot a^3 \cdot a^4 \ldots a^n\right)^{1 / n} \\ & \qquad=\left(a^{1+2+3+\ldots n}\right)^{1 / n}=a^{\left(\frac{n(n+1)}{2}\right)^{1 / n}}=a^{\frac{n+1}{2}} \\ & \text { Now, } \beta^2=a^{n+1} \\ & \text { and }\left(\frac{2 \alpha}{n+1}\right)^{n+1}=\left(\frac{2 a(n+1)}{2(n+1)}\right)^{n+1}=a^{n+1} \\ & \therefore \quad \beta^2=\left(\frac{2 \alpha}{n+1}\right)^{n+1}\end{aligned}$
So, locus of point $P(\alpha, \beta)$ is $y^2=\left(\frac{2 x}{n+1}\right)^{n+1}$
Hence, the correct answer is option: 'B'.