Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$.Draw a tangent at the point $p_1$ Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$. Draw a tangent at the point $p_1$ and let it meet the graph again at point $p_2$. Then draw a tangent at $p_2$ and so on . Show that , the ratio  $\frac{A(\bigtriangleup p_np_{n+1}p_{n+2})}{A( \bigtriangleup p_{n+1}p_{n+2}p_{n+3} )}$ is constant.
I don't think just taking the ratio of two triangles will help, something more need to be used.
 A: Expanding on my comment above, consider any point $P(a, a^5)$ other than the origin on the curve $f(x)$.  The slope of the tangent at $P$ is $5a^4$.  Hence the point of intersection is $Q(b, b^5)$, where $a \neq b$ and
$$5a^4 = \frac{b^5-a^5}{b-a} \implies b^3 + 2ab^2+3a^2b+4a^3 = 0$$
$$\implies r^3+2r^2+3r+4 = 0$$
where $r = b/a$.  Clearly as a cubic this has at least one real root, and it must be negative.  As the derivative $3r^2+4r+3$ has negative discriminant, it cannot have any more roots.  It is also easy to check $r < -1$.  Thus let $k = -r > 1$, and we have for any $P_n(x_n, x_n^5)$, the next point in sequence is $P_{n+1} (-kx_n, -k^5 x_n^5)$.  
Now the ratio of areas you seek is
$$\frac{A(\bigtriangleup p_np_{n+1}p_{n+2})}{A( \bigtriangleup p_{n+1}p_{n+2}p_{n+3} )} = \frac{\begin{vmatrix} x_n & x_n^5 & 1 \\ x_{n+1} & x_{n+1}^5 & 1 \\
x_{n+2} & x_{n+2}^5 & 1 \end{vmatrix}
}{\begin{vmatrix} x_{n+1} & x_{n+1}^5 & 1 \\ x_{n+2} & x_{n+2}^5 & 1 \\
x_{n+3} & x_{n+3}^5 & 1 \end{vmatrix}
} = \frac{\begin{vmatrix}
x_n & x_n^5 & 1 \\ 
-kx_n & -k^5x_n^5 & 1 \\ 
k^2x_n & k^{10}x_n^5 & 1 \\ 
\end{vmatrix}
}{\begin{vmatrix}
-kx_n & -k^5x_n^5 & 1 \\ 
k^2x_n & k^{10}x_n^5 & 1 \\ 
-k^3x_n & -k^{15}x_n^5 & 1 \\ 
\end{vmatrix}
} = \frac1{k^6
}$$
