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The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.

What I have attempted:

Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$

We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$

Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?

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  • $\begingroup$ What is the derivative $y'$? $\endgroup$
    – imranfat
    Aug 4, 2016 at 3:41
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    $\begingroup$ Notice affine transformation preserves tangency and ratio of areas, you can use an affine transform to map the ellipse to the unit circle. The smallest triangle $AOB$ will be mapped to a right angled isoceles triangle of side $\sqrt{2}$ and area $1$. This means the smallest triangle $AOB$ has area $ab$. $\endgroup$ Aug 4, 2016 at 4:00

2 Answers 2

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$A = \frac {a^2 b^2}{2xy}$ constrained by $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$

$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\ y + x y' = 0\\ y' = -\frac yx$

Differentiating the constraint.

$\frac x{a^2} + \frac {y y'}{b^2} = 0\\ y' = -\frac {b^2x}{a^2y}\\ \frac {b^2x}{a^2y} = \frac yx\\ b^2 x^2 = a^2 y^2$

Again from the constraint:

$b^2 x^2 + a^2 y^2 = a^2 b^2\\ 2a^2 y^2 = a^2 b^2\\ y^2 = \frac {b^2}{2}\\ y = \frac b{\sqrt {2}}\\ 2xy = ab\\ A = \frac {a^2b^2}{2xy} = ab$

alternate

$x = a \cos t\\ y = b \sin t\\ A = \frac {ab}{2} \csc t \sec t\\ \frac{dA}{dt} = \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\ \cot t = \tan t\\ t = \frac {\pi}{4}\\ A = ab$

That is much nicer.

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With the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\color{red}{ab}$, plain and simple.

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