Can't find mistake when calculating ratio between intersections of graphs

The function of the parabola is $$f(x) = x^2 - 1$$.
BA is tangent to the parabola at point A.
CA and BA intersect at point A.

Find the ratio between the dotted area and the dashed area.

Here's how I did it:

$$\text{Let the coordinates of A be } (x_0, y_0) \\ \text{Finding the coordinates of C:}$$ $$f(0) = 0^2 - 1 = -1 \rightarrow \boxed{C: (0, -1)}$$ $$\text{Finding the slope of CA to create the line equation:}$$ $$m_{CA} = \frac{y_0 +1}{x_0 - 0} = \frac{y_0 + 1}{x_0} \\ \text{Plugging the variables in the point-slope equation for CA:}$$ $$y_{CA} - -1 = \frac{y_0 + 1}{x_0}(x - 0) \rightarrow \boxed{y_{CA} =\frac{y_0 + 1}{x_0}x - 1}$$ $$\text{The slope of BA is equal to the derivative of the parabola at point A:}$$ $$f'(x_0) = 2x_0$$ $$\text{Plugging the variables in the point-slope equation for BA:}$$ $$y_{BA} - y_0 = 2x_0(x - x_o) \rightarrow \boxed{y_{BA} = 2x_0(x-x_0) + y_0}$$ $$\text{Calculating the area ratios:}$$ $$\frac{S_{\text{dotted}}}{S_{\text{dashed}}} = \frac{\int^{x_{0}}_0 [y_{CA} - f(x)]dx}{\int^{x_{0}}_0 [f(x) - y_{BA}]dx} = \frac{(\frac{y_0+1}{2x_0}\cdot x_0^2-x_0) - (\frac{x_0^3}{3}-x_0)}{(\frac{x_0^3}{3}-x_0)-(x_0^3 -2x_0^3 + y_0\cdot x_0)} \\ = \frac{\frac{3x_0(y_0+1)-6x_0 -2x_0^3+6x_0}{6}}{\frac{x_0^3-3x_0-3x_0^3+6x_0^3-3y_0x_0}{3}} = \frac{3x_0[3(y_0+1)-2x_0^2]}{6x_0[4x_0^2-3(y_0+1)]}$$

That's as far as I got. The correct answer turns out to be $$1/2$$. Now, it may look as though I have skipped many algebraic steps, but in my notes I have been very thorough, and only cut it short to spare time formatting. I have checked everywhere possible, and could not find what have I done wrong.

Do I have a mistake, or am I missing something?

• You made a typographical error $$y_{CA}-(-1)=\frac{y_0+1}{x_0}(x-0)\implies y_{CA}=\frac{y_0 + 1}{x_0}x-1$$ – Shubham Johri Jan 7 '19 at 16:33
• @ShubhamJohri Oops, fixed. – daedsidog Jan 7 '19 at 16:35

You didn't make a mistake. Just observe that $$y_0=x_0^2-1$$ as $$(x_0,y_0)$$ lies on $$y=x^2-1$$ and substitute in your answer to get $$1/2$$.