Question

$A_r;r=1,2,3,...,n$ are n points on the parabola $y^2=4x$ in the first quadrant. If $A_r=(x_r,y_r)$, where $x_1$, $x_2$, $x_3$, ..., $x_n$ are in GP and $x_1=1$, $x_2=2$, then $y_n$ is equal to

• A. $-2^{\frac{n+1}{2}}$
• B. $2^{n+1}$
• C. ${\left(\sqrt{2}\right)}^{n+1}$
• D. $2^{\frac{n}{2}}$

Answer 1

Answer: (C)

${\left(\sqrt{2}\right)}^{n+1}$

$x_r,y_r$ are points on the parabola $y^2=4x$ where $r=1,2,....,n$

$x_1=1$ and $x_2=2$

Since, $x_1,x_2,...,x_n$ are in G.P.

Therefore, $t_n=a{r}^{n-1}=x_n=x_1{\left(\frac{x_2}{x_1}\right)}^{n-1}=2^{n-1}$

Therefore, ${y_n}^2=4x_n=4\times 2^{n-1}=2^{n+1}$

$y_n={\left(\sqrt{2}\right)}^{n+1}$