MHB Dividing quarter of an ellipse

[suji]

Hi all,
How should I divide quarter of an ellipse into two equal havles? At what angle should I divide so that the 2 parts are equal?
Any hint is a privilege.
Thanks in advance
Regards
Suji

 

[MarkFL]

I would begin with an ellipse in standard form centered at the origin:

$$\left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1$$

Converting to polar coordinates, we may write:

$$r^2\left(\frac{\cos^2(\theta)}{a^2}+ \frac{\sin^2(\theta)}{b^2} \right)=1$$

Solving for $r^2$, we obtain:

$$r^2=\frac{(ab)^2}{a^2\sin^2(\theta)+b^2 \cos^2(\theta)}$$

Next, using the formula for area in polar coordinates, we obtain:

$$\int_0^{\beta}\frac{1}{a^2\sin^2(\theta)+b^2\cos^2(\theta)}\,d\theta= \int_{\beta}^{\frac{\pi}{2}}\frac{1}{a^2\sin^2( \theta)+b^2\cos^2( \theta)}\,d\theta$$

Applying the FTOC, we then obtain:

$$\left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_0^{\beta}=\left[\tan^{-1}\left(\frac{a}{b}\tan(\theta) \right) \right]_{\beta}^{\frac{\pi}{2}}$$

This gives us:

$$\tan^{-1}\left(\frac{a}{b}\tan(\beta) \right)=\frac{\pi}{4}$$

Taking the tangent of both sides:

$$\frac{a}{b}\tan(\beta)=1$$

Solve for $\beta$:

$$\beta=\tan^{-1}\left(\frac{b}{a} \right)$$

A much simpler technique would be to begin with the ellipse:

$$\left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=1$$

Now stretch the vertical axis by a factor of $$\frac{a}{b}$$ 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 $$\frac{b}{a}$$, and the dividing line is now:

$$y=\frac{b}{a}x$$

And we see the angle of inclination of this line is:

$$\beta=\tan^{-1}\left(\frac{b}{a} \right)$$

 

[suji]

Thanks a million sir MarkFL.
This helps me.:D