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I am stuck solving this geometry problem.

Given ellipse $x^2 + 4 y^2 = 32$ is bounding a rectangle whose height is the double of its width. At the apexes of rectangle are tangent lines to the ellipse. What is the ratio of rectangle area and obtained quadrangle?

What I could do is the following:

I transformed the equation to this form:

$$\frac{x^{2}}{32}+\frac{y^{2}}{8} = 1$$

and I got $a^{2}=32$ and $b^{2}=8$. The ratio of the rectangle if x and y are sides of rectangle is: $x:y = 2:1$ so $P=2y^2$. The equation of tangent line at $M(z,c)$ is

$$\frac{z\cdot x}{a^2} + \frac{c\cdot y}{b^2} = 1$$

I really have no clue what to do here to get this ratio. Any hints would be helpful! Thanks.

Rodrigo de Azevedo
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Filip Vlaisavljević
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1 Answers1

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Height is twice the width means if $x$ is the width and $y$ is the height then $x:y=1:2.$ But I think it will be easier if you say the width and height are $2z$ and $2c$ so that the rectangle has a vertex at $(z,c).$ Then the height-to-width ratio says $c=2z$.

Can you continue from there?

David K
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