Show that every tangent to the curve y = 1/2 sqrt(x-4x²)

Show that every tangent to the curve $y = \dfrac{1}{2}\sqrt{x-4x^{2}}$ meets the $y$-axis at a point $T_{1}$ that is equidistant from the origin $O$ and the point of tangency $P$.

Step 1: Domain and derivative

The radicand requires $x-4x^{2}=x(1-4x)\ge0$, so $x\in[0,\tfrac14]$. With $f(x)=\tfrac12(x-4x^{2})^{1/2}$, the chain rule gives

\[ y' = \frac{1-8x}{4\sqrt{x-4x^{2}}}. \tag{1} \]

Step 2: Tangent at an arbitrary point $P$

Let $P=(t,y_{0})$, $y_{0}=\tfrac12\sqrt{t-4t^{2}}$, $t\in(0,\tfrac14)$, with slope $k=\dfrac{1-8t}{4\sqrt{t-4t^{2}}}$. The tangent is $y-y_{0}=k(x-t)$.

Step 3: The $y$-intercept $T_{1}$

Setting $x=0$ and combining over the denominator $4\sqrt{t-4t^2}$ (the numerator $2(t-4t^2)-t(1-8t)$ collapses to $t$): \[y_{T_{1}} = y_{0}-kt = \frac{2(t-4t^{2})-t(1-8t)}{4\sqrt{t-4t^{2}}} = \frac{t}{4\sqrt{t-4t^{2}}}. \tag{2}\]

So $T_{1}=\left(0,\,\dfrac{t}{4\sqrt{t-4t^{2}}}\right)$.

Step 4: Compute $\overline{OT_{1}}$ and $\overline{PT_{1}}$

Since $t>0$ and $t-4t^{2}>0$, $y_{T_1}>0$ and

\[ \overline{OT_{1}} = y_{T_{1}} = \frac{t}{4\sqrt{t-4t^{2}}}. \tag{3} \]

With $P=(t,y_{0})$ and $T_{1}=(0,y_{T_1})$, and $y_{0}-y_{T_1}=kt$, \[\overline{PT_{1}}^{\,2} = t^{2}+(y_{0}-y_{T_1})^{2} = t^{2}(1+k^{2}). \tag{4}\]

Now $k^{2}=\dfrac{(1-8t)^{2}}{16(t-4t^{2})}$, and the numerator $16(t-4t^2)+(1-8t)^2 = 16t-64t^2+1-16t+64t^2=1$ collapses to $1$, so

\[ 1+k^{2} = \frac{1}{16(t-4t^{2})} \;\Longrightarrow\; \overline{PT_{1}}^{\,2} = \frac{t^{2}}{16(t-4t^{2})} \;\Longrightarrow\; \overline{PT_{1}} = \frac{t}{4\sqrt{t-4t^{2}}}. \tag{5} \]

Step 5: Verification

\[t=\tfrac{1}{10}\text{: } y_0=0.15,\ y_{T_1}=0.10206,\ \overline{OT_1}=0.10206,\ \overline{PT_1}=0.10206\ \checkmark\]