0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What do you mean by normal chord? $\endgroup$
    – user261263
    Nov 27 '16 at 11:12
  • 1
    $\begingroup$ @EugenCovaci That seems to be standard usage: it means the parabola's chord is normal (perpendicular) to the parabola's tangent line at the point of tangency . $\endgroup$
    – DonAntonio
    Nov 27 '16 at 11:14
0
$\begingroup$

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 $\theta=135°$.

$\endgroup$

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.