1
$\begingroup$

There is a analytic geometry problem:

In the ellipse $\frac{x^2}{4}+y^2=1$, segment $AB$ is a chord and $AB=\sqrt{3}$, find the maximum and minimum area of $\triangle AOB$.

enter image description here

My progress

Assume $A(x_1,y_1),B(x_2,y_2)$, we have $$ \left\{ \begin{array}{} (x_1-x_2)^2+(y_1-y_2)^2=3 \\ \frac{x_1^2}{4}+y_1^2=1 \\ \frac{x_2^2}{4}+y_2^2=1 \end{array} \right.$$ And $$ S_{\triangle AOB}=\frac{1}{2}(x_1y_2-x_2y_1)$$

After a complicated simplify, I got

$$ \frac{3}{4}x_1^2+\frac{3}{4}x_2^2-2x_1x_2-\frac{\sqrt{16-4x_1^2-4x_2^2+x_1^2x_2^2}}{2}=1$$

and the target $\frac{1}{2}(x_1y_2-x_2y_1)$ turned to

$$ \frac{\sqrt{4x_1^2-x_1^2x_2^2}-\sqrt{4x_2^2-x_1^2x_2^2}}{2}$$

It likely is a conditional extremum problem and I don't know what to do next.

$\endgroup$
  • $\begingroup$ Have you tried using the parametric form of the ellipse? $\endgroup$ – David Quinn Apr 24 '16 at 10:30
2
$\begingroup$

It likely is a conditional extremum problem and I don't know what to do next.

I don't know what to do next either.

So, let us take another approach.

If the line $AB$ is parallel to $y$-axis, we can easily see that the area of the triangle is $\sqrt 3/2$.

If the line $AB$ is not parallel to $y$-axis, we can set $AB : y=ax+b$.

Then, $$\frac{x^2}{4}+(ax+b)^2=1\iff (4a^2+1)x^2+8abx+4b^2-4=0$$ Let $\alpha,\beta$ be the $x$ coordinates of $A,B$. Then, $$\alpha+\beta=\frac{-8ab}{4a^2+1},\quad\alpha\beta=\frac{4b^2-4}{4a^2+1}$$

Now, we can write $$\sqrt 3=AB=\sqrt{(\alpha-\beta)^2+((a\alpha+b)-(a\beta+b))^2}$$ Squaring the both sides gives $$3={(a^2+1)((\alpha+\beta)^2-4\alpha\beta)}={(a^2+1)\left(\left(\frac{-8ab}{4a^2+1}\right)^2-4\frac{4b^2-4}{4a^2+1}\right)}$$ and so $$b^2=\frac{16a^4+56a^2+13}{16(a^2+1)}$$

Let $S$ be the area of the triangle. Then, $$S^2=\left(\frac{\sqrt 3}{2}\cdot\frac{|a\cdot 0-0+b|}{\sqrt{a^2+1}}\right)^2=\frac{3b^2}{4(a^2+1)}=\frac{3(16a^4+56a^2+13)}{64(a^2+1)^2}:=f(a)$$ $$f'(a)=\frac{9a(5-4a^2)}{16(a^2+1)^3}=0\iff a=0,\pm\frac{\sqrt 5}{2}$$ Hence, the maximum of $f(a)$ is $f(\pm\sqrt 5/2)=1$, and the minimum of $f(a)$ is $f(0)=39/64$.

Therefore, the maximum area of the triangle is $\color{red}{1}$, and the minimum area is $\color{red}{\sqrt{39}/8}$.

$\endgroup$

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.