Prove that every tangent of a function cuts $y$ axis at a point that is at equal distance from (0,0) and touching point

Question

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.

Answer 1

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:

$$(y-y_0)=y'_0(x-x_0)$$

where:

$$y'_0=\frac{1-8x_0}{4\sqrt{x_0-4x_0^2}}=\frac{1-8x_0}{8y_0}$$

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

$$y_1-y_0=\frac{8x_0^2-x_0}{8y_0}\tag{2}$$

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

$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$

$$(y_1-y_0)^2+x_0^2=y_1^2$$

$$-2y_0y_1+y_0^2+x_0^2=0$$

$$x_0^2+y_0^2=2y_0y_1\tag{3}$$

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

$$x_0^2+y_0^2=2y_0(y_0+\frac{8x_0^2-x_0}{8y_0})$$

$$x_0^2=y_0^2+\frac14(8x_0^2-x_0)$$

$$4x_0^2=4y_0^2+8x_0^2-x_0$$

$$x_0-4x_0^2=4y_0^2$$

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

Answer 2

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
$p = \sqrt{(p - y(a))^2 + a^2}.$

Answer 3

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.