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Q. Let A be the region in the xy-plane given by

$$A=\{(x,y): x=u+v, y=v, u^2+v^2≤1 \} $$

Derive the length of the longest line segment that can be enclosed inside region A.

My attempt: I found the equation of the region A:

$$x^2+2y^2-2xy-1≤0$$

This is the region enclosed by an ellipse. So the major axis must be the longest line segment. But, I am not able to find the major axis. Please help me out.

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2 Answers 2

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Consider the map $f: (u, v) \mapsto (u+v, v)$. This linear map converts from $uv$-coordinates to $xy$-coordinates. If $D$ is the unit circle in the $uv$-plane, then $A = f(D)$ is the ellipse we are looking for. Now the length of the major axis of the eclipse is exactly twice the largest possible length of the image of a unit vector under $f$, so the length you are looking for is exactly $2\|f\|_2$.

Do you know how to find $\|f\|_2$ or should I continue?

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  • $\begingroup$ Thank you very much for your help.. $\endgroup$
    – Utkarsh
    Commented Jun 1, 2016 at 12:30
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Eliminate u and v.

The Inequality is:

$$(x-y)²+(y)²≤1$$

$$x²+2y²-2xy-1≤0$$

If we let

$$S(x,y)=x²+2y²-2xy-1=0$$

Then we have $h²-ab<0$ indicating that S is an ellipse. So we need to find the major and minor axis of the ellipse. First: The centre of ellipse is $ (0,0) $because,$ (0,0)$ is the solution of $$\frac{\partial S}{\partial x} =0$$ $$\frac{\partial S}{\partial y} =0$$ Now put $x=rcos(t)$ and $y=r sin(t)$ in $S$ to get, after simplification

$$r²=\frac{1}{3-(cos(2t) + sin(2t))}$$ $$\frac{2}{3+\sqrt5}≤r²≤ \frac{2}{3-\sqrt5}$$

Thus the semi major and semi minor axis of ellipse are $$a²=\frac{2}{3-\sqrt5}$$

$$b²=\frac{2}{3+\sqrt5}$$

So longest line segment has length equal to $2a$ which is $$l(max)=2 \sqrt{\frac{2}{3-\sqrt5}}=\sqrt{2(3+\sqrt 5)}$$

Note that the above method yields the lengths of axis of ellipse or hyperbola whose centre is at $(0,0)$.

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