- #26

fresh_42

Mentor

- 15,421

- 13,459

Yes, but I thought of a geometric proof which would have been shorter (therefore the picture).Do you want me to derive the equation of director circle?

Let ##y=mx+c## be a tangent to the ellipse ##\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1##, so on solving these two we get,

\begin{align*}

b^2x^2+a^2(mx+c)^2&=a^2b^2\\

(b^2+a^2m^2)x^2+(2a^2mc)x+a^2(c^2-b^2)&=0

\end{align*}

For ##y=mx+c## to be a tangent to the ellipse ##\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1##, discriminant of the above quadratic equation must be zero (to ensure that the line cuts the ellipse at only one point),

\begin{align*}

4a^4m^2c^2-4a^2(b^2+a^2m^2)(c^2-b^2)&=0\\

4a^2(a^2m^2c^2-b^2c^2+b^4-a^2m^2c^2+a^2b^2m^2)&=0\\

4a^2b^2(-c^2+b^2+a^2m^2)&=0\\

c^2=a^2m^2+b^2\\

c=\pm \sqrt{a^2m^2+b^2}

\end{align*}

So, the equation of tangent to the ellipse is $$y=mx \pm \sqrt{a^2m^2+b^2}$$

Let the locus of the director circle is ##P(h,k)##

Now, as the tangent passes through ##P(h,k)##, we can write,

\begin{align*}

&y=mx \pm \sqrt{a^2m^2+b^2}\\

&k=mh \pm \sqrt{a^2m^2+b^2}\\

&k-mh= \pm \sqrt{a^2m^2+b^2}\\

&k^2+m^2h^2-2khm=a^2m^2+b^2\\

&(a^2-h^2)m^2+(2hk)m+(b^2-k^2)=0

\end{align*}

This is a quadratic equation in ##m## so its roots represents slope of tangents passing through ##P(h,k)## and by the definition of director the circle, they are perpendicular,

\begin{align*}

m_1m_2&=-1\\

\dfrac{b^2-k^2}{a^2-h^2}&=-1\\

h^2+k^2&=a^2+b^2\\

x^2+y^2&=a^2+b^2

\end{align*}

which is the locus of ##P(h,k)##

And yes, we don't get the entire circle as the ellipse is rotating only in the first quadrant, we get an arc of the circle ##y=\sqrt{5-x^2}## from ##x=1## to ##x=2## because when the major axis of the ellipse is horizontal, the ##x## coordinate of the center is ##x=2## (maximum case as on rotating the ##x## coordinate of the center decreases) and when the major axis of ellipse is vertical, the ##x## coordinate of the center is ##x=1## (the other maximum case as the ellipse cannot cross the coordinate axes, the distance from the center of the ellipse to the y-axis is equal to the length of semi-minor axis of the ellipse)