enter image description here

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?

  • 1
    $\begingroup$ You made a typographical error $$y_{CA}-(-1)=\frac{y_0+1}{x_0}(x-0)\implies y_{CA}=\frac{y_0 + 1}{x_0}x-1$$ $\endgroup$ Jan 7, 2019 at 16:33
  • $\begingroup$ @ShubhamJohri Oops, fixed. $\endgroup$
    – daedsidog
    Jan 7, 2019 at 16:35

1 Answer 1


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 $1/2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.