Question:
Tangent at a point $C_1$ on the curve $y=x^3$ meets the curve again at $C_2$. The tangent at $C_2$ meets the curve at $C_3$, and so on, so that the abscissa of $C_1,C_2,C_3,….,C_n$ form a G.P. Find the ratio of area of triangle $C_1C_2C_3$ to the area of triangle $C_2C_3C_4$.
Attempt:
By equating two points where the tangents meet got a relation between their x coordinates as :
$X_2=(-2)X_1$
$X_3=(-2)X_2=(-2)^2X_1$
$X_4=(-2)^3X_1$
$X_5=(-2)^4X_1$
And so on…
How to proceed now also how to use this information about abscissa?
Best Answer
Well, once you know the $x$-coordinates you can compute the $y$-coordinates, finally you can use the area formula for a triangle $ABC$ in terms of its coordinates:
$$area = \left|\frac {A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}2\right|.$$
Here $A_x$ and $A_y$ are the $x$- and $y$-coordinates of the point $A$. (See for example this page.)
Alternatively, you could use that area = 1/2 base * height, and the fact that your triangles have a common base in $c_2c_3$ - thus the ratio of areas will just be the ratio of heights of $c_1$ and $c_4$ with respect to the common base.
(To compute the altitudes, translate things so that $C_2 = (c_2, f(c_2))$ is at the origin, then find an orthogonal to the translate of $C_3$, finally take its dot product with the translate of $C_4$.)