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Dividing quarter of an ellipse
- Thread starter suji
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I would begin with an ellipse in standard form centered at the origin:
\(\displaystyle \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1\)
Converting to polar coordinates, we may write:
\(\displaystyle r^2\left(\frac{\cos^2(\theta)}{a^2}+ \frac{\sin^2(\theta)}{b^2} \right)=1\)
Solving for $r^2$, we obtain:
\(\displaystyle r^2=\frac{(ab)^2}{a^2\sin^2(\theta)+b^2 \cos^2(\theta)}\)
Next, using the formula for area in polar coordinates, we obtain:
\(\displaystyle \int_0^{\beta}\frac{1}{a^2\sin^2(\theta)+b^2\cos^2(\theta)}\,d\theta= \int_{\beta}^{\frac{\pi}{2}}\frac{1}{a^2\sin^2( \theta)+b^2\cos^2( \theta)}\,d\theta\)
Applying the FTOC, we then obtain:
\(\displaystyle \left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_0^{\beta}=\left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_{\beta}^{\frac{\pi}{2}}\)
This gives us:
\(\displaystyle \tan^{-1}\left(\frac{a}{b}\tan(\beta) \right)=\frac{\pi}{4}\)
Taking the tangent of both sides:
\(\displaystyle \frac{a}{b}\tan(\beta)=1\)
Solve for $\beta$:
\(\displaystyle \beta=\tan^{-1}\left(\frac{b}{a} \right)\)
A much simpler technique would be to begin with the ellipse:
\(\displaystyle \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1\)
Now stretch the vertical axis by a factor of \(\displaystyle \frac{a}{b}\) such that the ellipse now becomes a circle of radius $a$. We know the line $y=x$ will divide the first quadrant area of the circle into two equal halves.
Now shrink the vertical axis back to where it started, by a factor \(\displaystyle \frac{b}{a}\), and the dividing line is now:
\(\displaystyle y=\frac{b}{a}x\)
And we see the angle of inclination of this line is:
\(\displaystyle \beta=\tan^{-1}\left(\frac{b}{a} \right)\)
\(\displaystyle \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1\)
Converting to polar coordinates, we may write:
\(\displaystyle r^2\left(\frac{\cos^2(\theta)}{a^2}+ \frac{\sin^2(\theta)}{b^2} \right)=1\)
Solving for $r^2$, we obtain:
\(\displaystyle r^2=\frac{(ab)^2}{a^2\sin^2(\theta)+b^2 \cos^2(\theta)}\)
Next, using the formula for area in polar coordinates, we obtain:
\(\displaystyle \int_0^{\beta}\frac{1}{a^2\sin^2(\theta)+b^2\cos^2(\theta)}\,d\theta= \int_{\beta}^{\frac{\pi}{2}}\frac{1}{a^2\sin^2( \theta)+b^2\cos^2( \theta)}\,d\theta\)
Applying the FTOC, we then obtain:
\(\displaystyle \left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_0^{\beta}=\left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_{\beta}^{\frac{\pi}{2}}\)
This gives us:
\(\displaystyle \tan^{-1}\left(\frac{a}{b}\tan(\beta) \right)=\frac{\pi}{4}\)
Taking the tangent of both sides:
\(\displaystyle \frac{a}{b}\tan(\beta)=1\)
Solve for $\beta$:
\(\displaystyle \beta=\tan^{-1}\left(\frac{b}{a} \right)\)
A much simpler technique would be to begin with the ellipse:
\(\displaystyle \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1\)
Now stretch the vertical axis by a factor of \(\displaystyle \frac{a}{b}\) such that the ellipse now becomes a circle of radius $a$. We know the line $y=x$ will divide the first quadrant area of the circle into two equal halves.
Now shrink the vertical axis back to where it started, by a factor \(\displaystyle \frac{b}{a}\), and the dividing line is now:
\(\displaystyle y=\frac{b}{a}x\)
And we see the angle of inclination of this line is:
\(\displaystyle \beta=\tan^{-1}\left(\frac{b}{a} \right)\)
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