Basic Examples On Ellipses Set-3

Go back to  'Ellipse'

Example - 6

Show that the sum of the reciprocals of the squares of any two diameters of an ellipse which are at right angles to one another is a constant.

Solution: By a diameter of an ellipse, we mean any chord which passes through its centre.

Let \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) be the ellipse and let AB and CD be any two diameters of the ellipse perpendicular to each other.

Since AB and CD are diameters, we can assume AO = OB = r1, and CO = OD = r2. Also, if the slope of AB is given by \(\theta \) than that of CD is obviously \(\begin{align}\frac{\pi }{2} + \theta .\end{align}\)

Thus, we get the co-ordinates of A, B, C and D as

\[\begin{align}&A,\,\,B \equiv \pm ({r_1}\cos {\theta _1},\,{r_1}\sin \theta ){\rm{ }}\qquad  C,D \equiv \pm \left( {{r_2}\cos \left( {\frac{\pi }{2} + \theta } \right),\,\,{r_2}\sin \left( {\frac{\pi }{2} + \theta } \right)} \right)\\&{\rm{\; }}\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad= \pm ( - {r_2}\sin \theta ,\,\,{r_2}\cos \theta )\end{align}\]

These coordinates must satisfy the equation of the ellipse; we therefore obtain :


\[\begin{align}&\frac{{r_1^2{{\cos }^2}\theta }}{{{a^2}}} + \frac{{r_1^2{{\sin }^2}\theta }}{{{b^2}}} = 1\;and\;\frac{{r_2^2{{\sin }^2}\theta }}{{{a^2}}} + \frac{{r_2^2{{\cos }^2}\theta }}{{{b^2}}} = 1\\&\Rightarrow \qquad \frac{1}{{r_1^2}} = \frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}\;and\;\frac{1}{{r_2^2}} = \frac{{{{\sin }^2}\theta }}{{{a^2}}} + \frac{{{{\cos }^2}\theta }}{{{b^2}}}\end{align}\]

Adding these two relations, we have

\[\begin{align}&\frac{1}{{r_1^2}} + \frac{1}{{r_2^2}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}\\& \Rightarrow \qquad \frac{1}{{{{(2{r_1})}^2}}} + \frac{1}{{{{(2{r_2})}^2}}} = \frac{1}{{4{a^2}}} + \frac{1}{{4{b^2}}}\\&\Rightarrow\qquad \frac{1}{{A{B^2}}} + \frac{1}{{C{D^2}}} = \frac{1}{{4{a^2}}} + \frac{1}{{4{b^2}}}\left( {a{\rm{ \;}}constant} \right)\end{align}\]

This proves the assertion stated in the question.

AUXILIARY CIRCLE

Example – 7

Let \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) be an ellipse. Assume a > b. A circle is described on the major axis of this ellipse as diameter. From any point P on this circle, a perpendicular PQ is dropped onto the major axis of the ellipse. Show that PQ will always be divided in a fixed ratio by the ellipse.

Solution:

As is evident, the radius of this circle, called the auxiliary circle of the ellipse, is a, so that its equation is

\[{x^2} + {y^2} = {a^2}\]

Now any point P on this circle can be taken in parametric form as \(P \equiv (a\cos \theta ,\,a\sin \theta )\) where \(\theta \) is the angle that OP makes with the horizontal.

To evaluate the y-coordinate of R, we substitute \(x = a\cos \theta \) in the equation of the ellipse:

\[\begin{align}&\frac{{{a^2}{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\\&\Rightarrow  y = b\sin \theta\end{align}\]

Thus, R is the point \((a\cos \theta ,\,b\sin \theta )\) while Q is simply \((a\cos \theta ,\,0).\) We now see that

\[\frac{{PR}}{{RQ}} = \frac{{a\sin \theta - b\sin \theta }}{{b\sin \theta }} = \frac{a}{b} - 1\]

which is independent of \(\theta \) proving the stated assertion.

There is one significant fact that we can learn about the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.\) We obtained the coordinates of R as\((a\cos \theta ,\,b\sin \theta ).\) This tells us that an alternative way to specify an ellipse is in terms of a parameter \(\theta \) :

\[\boxed{x = a\cos \theta ,\,\,y = b\sin \theta }\]

This is referred to as the parametric form of the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.\)\(\theta \) is called the eccentric angle of the point \((x,\,y) \equiv (a\cos \theta ,\,\,b\sin \theta ).\) It is important to note that \(\theta \) is not the angle that \((a\cos \theta ,\,\,b\sin \theta )\) makes with the horizontal; it is the angle which the corresponding point on the auxiliary circle makes with the horizontal. Depending on what value \(\theta \) takes in the range\([0,\,2\pi )\)\(({\rm{or ( - }}\pi {\rm{, }}\pi ]),\)the parametric form \((a\cos \theta ,\,b\sin \theta )\) gives us different points on the circumference of the ellipse.

The point \((a\cos \theta ,\,b\sin \theta )\) in sometimes simply referred to as the point \(\theta \)

Example - 8

What are the eccentric angles of the extremities of the latus-recta in the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,?\)

Solution: As discussed earlier, the co-ordinates of the end-points of the latus-recta are \(\left( { \pm ae,\,\, \pm \frac{{{b^2}}}{a}} \right).\) Thus, if we assume the required eccentric angle to be \(\phi ,\)we have

\[\begin{align}&a\cos \phi = \pm ae\\&b\sin \phi = \pm \frac{{{b^2}}}{a}\end{align}\]

This gives us four values of \(\phi ,\) given by

\[\tan \phi  =  \pm \frac{b}{{ae}}\]

corresponding to the four extremities of the two latus-recta: