Minimum area of a triangle formed by a normal chord and tangents at its extremities.

Question

If a chord is normal to the parabola $y^2=4ax$ and is inclined at an angle $\theta$ to the positive $x-axis$, then find the value of $\theta$ for which the area of the triangle, formed by the chord and the tangents at its extremities, is minimum.

Answer 1

Given the symmetry of the problem, we can consider the case $a>0$ without loss of generality, so we have a parabola $4ax=y^2$ passing through the origin $O(0,0)$, with x-axis as symmetry axis.

Let's take two points $U,V\ne O(0,0)$ of the parabola, so we have: $U({u^2\over4a},u)$ and $V({v^2\over4a},v)$, because $x={y^2\over4a}$.

Now we can get the tangent lines to the parabola in the points $U$ and $V$, using a general formula.

If $x=Ay^2+By+C$ is a parabola with horizontal symmetry axis, then we can get the tangent line in any point $P(x_p,y_p)$, belonging to the parabola, by the following formula:

$${x+x_p\over2}=Ayy_p+B{y+y_p\over2}+C$$

In this case $B=C=0$ and $A={1\over4a}$, thus we obtain the tangent lines $t_U$ and $t_V$:

\begin{equation} {x+\frac{u^2}{4a}\over2}=\frac{1}{4a}yu\Longleftrightarrow t_U: y=\frac{2ax}{u}+{u\over 2}\\ {x+\frac{v^2}{4a}\over2}=\frac{1}{4a}yv\Longleftrightarrow t_V: y=\frac{2ax}{v}+{v\over 2} \end{equation}

Now we can search the point $T$ of intersection between $t_U$ and $t_V$:

\begin{equation} \begin{cases} y=\frac{2ax}{u}+{u\over 2}\\ y=\frac{2ax}{v}+{v\over 2} \end{cases} \end{equation}

It is easy to solve this system so we get $T({uv\over4a},{u+v\over2})$. These three points define a the triangle $UTV$. This must be a right triangle because the chord $\overline{UV}$ is normal, as the problem requires, to the parabola; we can suppose that the chord is normal to the tangent line $t_U$ at the point $U$ (it would be the same for $V$). To enforce this condition, between the chord and the line $t_U$, we must evaluate the angular coefficient $m_{\overline{UV}}$:

$$m_{\overline{UV}}=\frac{u-v}{{1\over4a}(u^2-v^2)}=\frac{4a}{u+v}$$

The angular coefficient of $t_U$ is $\frac{2a}{u}$, hence:

$$m_{\overline{UV}}m_{t_U}=-1$$

and we get the perpendicularity condition:

\begin{equation} {8a^2=-u(u+v)} \end{equation}

Now we can calculate the distances $UT$ and $UV$, i.d. the two catheti:

$$UT=\sqrt{\left({u^2\over4a}-{uv\over4a}\right)^2+\left(u-{u\over2}-{v\over2}\right)^2}={|u-v|\over2}\sqrt{{u^2\over4a^2}+1}\\UV=\sqrt{\left({u^2\over4a}-{v^2\over4a}\right)^2+\left(u-v\right)^2}=|u-v|\sqrt{{(u+v)^2\over16a^2}+1}$$

The area of the triangle will be:

$$\mathcal{A}={UT\cdot UV\over 2}={(u-v)^2\over4}\sqrt{\left({u^2\over4a^2}+1\right)\left({(u+v)^2\over16a^2}+1\right)}={(u^2+4a^2)^2\over4au^2}$$

where the perpendicularity condition has been used to obtain the result.

To find the triangle with minimum area, we must derive the function $\mathcal{A}(u)$ and equal it to zero:

$$\mathcal{A'}(u)={2(u^2+4a^2)2u4au^2-8au(u^2+4a^2)^2\over16a^2u^4}={(u^2+4a^2)(u^2-4a^2)\over2au}=0$$

so we have $u=\pm 2a$. The solution is $u=2a$ as it can be easily seen.

If $u=2a$ then:

$$U(a,2a),V(9a,-6a),m_{\overline{UV}}=-1$$

but $m_{\overline{UV}}=\tan\theta$, so $\tan\theta=-1$ and $\theta=135°$.