If a function y = $\frac{1}{2}$$\sqrt{x-4x^2}$ is given, how would one prove that every tangent of the function cuts $y$ axis in a point that is at equal distance from point $(0, 0)$ and the point at which the tangent touches the function?Finding a derivative for every single point, constructing a tangent, finding a point where $x=0$ and then comparing given $y$ to the distance between $y$ and touching point is clearly not feasible.

  • $\begingroup$ Why can't it be done? $\endgroup$ Jan 19, 2019 at 10:32
  • $\begingroup$ Because there are infinitely many points on x axis? $\endgroup$
    – JoeDough
    Jan 19, 2019 at 10:41
  • $\begingroup$ For what reason? $\endgroup$
    – JoeDough
    Jan 19, 2019 at 10:50
  • $\begingroup$ You have received 3 answers, consider accepting one of them $\endgroup$
    – Saša
    Jan 21, 2019 at 9:12

3 Answers 3


Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:

$$y_0^2=\frac 14(x_0-4x_0^2)\tag{1}$$

Equation of tangent passing through $T$ is:




Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:


You have to show that: $TP=OP$ or $TP^2=OP^2$:





Combine (2) and (3) and you get:





...which is always true, according to (1)


The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).

So, tra la, you want to show
$p = \sqrt{(p - y(a))^2 + a^2}.$


This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.


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