Question

In an ellipse the distance between the center and any normal doesn’t exceed the difference between the semi axes of the ellipse.
$d \leqslant \left| {a - b} \right|$

Answer 1

Hint: Given that there is an ellipse and we consider that its center is the origin. We have to prove that the distance between the center and the normal to the ellipse is equal to or does not exceed the difference of the semi-major axis and semi-minor axis. To prove this we have to know about the normal equation of an ellipse and the general equation of an ellipse.

Complete step by step answer:
We know that the equation of an ellipse is given by:
$ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$

Its center at $({x_1},{y_1}) = (0,0)$
The equation of a the normal to the ellipse is given by:
$ \Rightarrow \dfrac{{{a^2}x}}{{{x_2}}} - \dfrac{{{b^2}y}}{{{y_2}}} = {a^2} - {b^2}$
Where the coordinates of $({x_2},{y_2}) = (a\cos \theta ,b\sin \theta )$
Substitute these $({x_2},{y_2})$ co-ordinates in the equation of normal of the ellipse, as given below:
$ \Rightarrow \dfrac{{{a^2}x}}{{a\cos \theta }} - \dfrac{{{b^2}y}}{{b\sin \theta }} = {a^2} - {b^2}$
$ \Rightarrow ax\sec \theta - by\cos ec\theta = {a^2} - {b^2}$
$\therefore $ The normal equation of an ellipse is $ax\sec \theta - by\cos ec\theta = {a^2} - {b^2}$.
Now given that the distance between the center and any normal doesn’t exceed the difference between the semi axes of the ellipse, we have to prove this.
Here the length of the major axis = 2a
Hence the length of the semi-major axis = a
The length of the minor axis = 2b
Hence the length of the semi-minor axis = b
The distance from the center to the normal is given by:
$ \Rightarrow \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Here $({x_1},{y_1}) = (0,0)$, and the coefficients a, b and c are the coefficients of the normal of the ellipse.
The normal equation of the ellipse is : $ax\sec \theta - by\cos ec\theta = {a^2} - {b^2}$, which can be re-written as:
$ \Rightarrow (a\sec \theta )x - (b\cos ec\theta )y - ({a^2} - {b^2}) = 0$
Now calculating the distance of normal from the center, as given below:
\[ \Rightarrow \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right| = \left| {\dfrac{{a(0) + b(0) - ({a^2} - {b^2})}}{{\sqrt {{{(a\sec \theta )}^2} + {{(b\cos ec\theta )}^2}} }}} \right|\]
$ \Rightarrow \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2}{{\sec }^2}\theta + {b^2}\cos e{c^2}\theta } }}} \right|$
We have to find the maximum of the distance, hence finding the minimum value of the denominator, as given below:
Consider ${a^2}{\sec ^2}\theta + {b^2}\cos e{c^2}\theta $, finding the minimum of it.
Let $x = {a^2}{\sec ^2}\theta + {b^2}\cos e{c^2}\theta $
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = 2{a^2}\sec \theta (\sec \theta \tan \theta ) + 2{b^2}\cos ec\theta ( - \cos ec\theta \cot \theta )$
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = 2{a^2}{\sec ^2}\theta \tan \theta - 2{b^2}\cos e{c^2}\theta \cot \theta $
To find the minimum value, $\dfrac{{dx}}{{d\theta }} = 0$
$ \Rightarrow 2{a^2}{\sec ^2}\theta \tan \theta - 2{b^2}\cos e{c^2}\theta \cot \theta = 0$
$ \Rightarrow 2{a^2}{\sec ^2}\theta \tan \theta = 2{b^2}\cos e{c^2}\theta \cot \theta $
Dividing the equation with 2 on both sides, as given below:
$ \Rightarrow \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{\cos e{c^2}\theta \cot \theta }}{{{{\sec }^2}\theta \tan \theta }}$
$ \Rightarrow \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{{{\cos }^4}\theta }}{{{{\sin }^4}\theta }}$
$ \Rightarrow \dfrac{a}{b} = \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}$
$\therefore \dfrac{a}{b} = {\cot ^2}\theta $
Hence $\dfrac{b}{a} = {\tan ^2}\theta $
Now substituting these values in the distance of normal from the center expression, as given below:
$ \Rightarrow \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2}{{\sec }^2}\theta + {b^2}\cos e{c^2}\theta } }}} \right| = \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2}(1 + {{\tan }^2}\theta ) + {b^2}(1 + {{\cot }^2}\theta )} }}} \right|$
\[ \Rightarrow \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2}\left( {1 + \dfrac{b}{a}} \right) + {b^2}\left( {1 + \dfrac{a}{b}} \right)} }}} \right|\]
$\because $We obtained that $\dfrac{b}{a} = {\tan ^2}\theta $ and $\dfrac{a}{b} = {\cot ^2}\theta $
\[ \Rightarrow \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2}\left( {\dfrac{{a + b}}{a}} \right) + {b^2}\left( {\dfrac{{b + a}}{b}} \right)} }}} \right|\]
\[ \Rightarrow \left| {\dfrac{{{a^2} - {b^2}}}{{\sqrt {a(a + b) + b(b + a)} }}} \right|\]
$\because $ We know that ${a^2} - {b^2} = (a + b)(a - b)$, substituting in the expression:
\[ \Rightarrow \left| {\dfrac{{(a + b)(a - b)}}{{\sqrt {(a + b)(a + b)} }}} \right|\]
\[ \Rightarrow \left| {\dfrac{{(a + b)(a - b)}}{{(a + b)}}} \right|\]
\[ \Rightarrow \left| {a - b} \right|\]
Hence proved.

The distance between the center and any normal doesn’t exceed the difference between the semi axes of the ellipse, which is given by \[\left| {a - b} \right|\].

Note: Here while finding the distance between the center of the ellipse and its normal, it is involved with modulus, because the modulus means that anything negative inside makes it positive outside. The most crucial thing to understand here is while finding the maximum value of the distance, we have to find the minimum value of the denominator, as it makes the whole value maximum. Here finding the minimum value of the denominator by differentiating it.