# Finding the area enclosed by the locus of the vertex of the rectangle at which the normals meet.

#### WMDhamnekar Let a and b be the lengths of the semimajor and semiminor axes of an ellipse respectively.

Draw a rectangle whose two sides are tangent to the ellipse and the other two are normal to the ellipse.

Now how to find the area enclosed by the locus of the vertex of the rectangle at which the normals meet.

We are required to find the locus of the point (h,k) from which two mutually perpendicular lines can be drawn that are normal to the ellipse.

Normal to the ellipse at the point $(a *\cos{\theta},b*\sin{\theta})$ is given by $a*x\sec{\theta}- b*y\csc{\theta}=a^2-b^2$ and slope of this normal is given by $m=\frac{a}{b} \tan{\theta}$

Now how to proceed further?
Note:- Graphical visualisation is appreciated.

#### WMDhamnekar

Hello,
If we put x=h, y=k, how to eliminate $\theta$ so that we can get equation in m?

#### WMDhamnekar

Hello,
After doing some computations and receiving guidance from math scholars from internet, I got the following equation.

$b^2h^2m^4-2*b^2h*km^3+(a^2h^2+b^2k^2-a^4e^4)*m^2-2a^2hkm+a^2k^2=0$ which is a quartic equation in m. The term $2a^2e^2m\sqrt{a^2+b^2m^2}*(k-mh)$ is of no use.(why?)
Now there are four roots to this quartic equation , namely $m_1, m_2, m_3, m_4$. For our locus, let us assume $m_1,m_2$ are the slopes of mutually perpendicular normals.

Now we know that the sums and products of roots of quartic equations are

$m_1+m_2+m_3+m_4=\frac{2k}{h}$

$m_1m_2+m_1m_3+m_1m_4+m_2m_3+m_2m_4+m_3m_4=\frac{a^2h^2+b^2k^2-a^4e^4}{b^2h^2}$

$m_1m_2m_3+m_1m_2m_4+m_1m_3m_4+m_2m_3m_4=\frac{2a^2k}{b^2h}$

$m_1m_2m_3m_4=\frac{a^2k^2}{b^2h^2}$

and $m_1m_2=-1$ and $e^4=(-\frac{b^2-a^2}{a^2})^2$

Now how to eliminate m to find the locus in terms of a, b,h,k ?

Answer provided to me 1) The graph of the locus of points (h,k) from which two mutually perpendicular lines can be drawn which are normals to the ellipse is as below. and the required area is $\pi(a-b)^2$

Now how to proceed further?

#### WMDhamnekar

Hello,

I am able to find out all the four roots to this quartic equation. $m_1$ is Here. Now $m_2=-\frac{1}{m_1}, m_3=\frac{ak}{bh},m_4=-\frac{ak}{bh}$

But if we plug in these values of roots in the following equation, $$m_1m_2m_3+m_1m_2m_4+m_1m_3m_4+m_2m_3m_4=\frac{2a^2k}{b^2h}$$, it becomes $$-\frac{2a^2k^3}{b^2h^3}=\frac{2a^2k}{b^2h}$$.

What does this mean? Does it mean $-h=k$ and/or vice versa. Would any member here explain me this discrepancy?

I am also able to find out $$(a^2-b^2)^2=(a^2+b^2)(h^2+k^2)$$ which will be helpful in computing the following equation.

$$(a^2+b^2)(h^2+k^2)(a^2k^2+b^2h^2)^2=(a^2-b^2)^2 (a^2k^2-b^2h^2)^2$$. This is the required locus.

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#### WMDhamnekar

Hello,

I solved this problem correctly and satisfactorily.

#### WMDhamnekar

Hello,

I finally got $r^2=\frac{(a^2-b^2)^2}{(a^2+b^2)}\left(\frac{a^2*\sin^2{\theta}-b^2*\cos^2{\theta}}{a^2*\sin^2{\theta}+b^2*\cos^2{\theta}}\right)^2$ by transforming the locus equation mentioned in #4 of this thread to polar coordinates. $h=r\cos{\theta}, k=r\sin{\theta}$

But when i plotted this equation in Wolframalpha computational intelligence, it is unable to plot this equation. So how to plot it in any graphing application or in www.wolframalpha.com?