Question

Prove that in an ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , the distance between the center and any normal doesn't exceed the difference between the semi axes of the ellipse.
$ d\le \ |a-b| $

Answer 1

Hint:The equation of the normal to the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , at
a point $ ({{x}_{1}},{{y}_{1}}) $ on it, is given by: $ \dfrac{{{a}^{2}}x}{{{x}_{1}}}-
\dfrac{{{b}^{2}}y}{{{y}_{2}}}={{a}^{2}}-{{b}^{2}} $ .

The co-ordinates any point on the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ are
of the form $ \left( a\cos \theta ,\ b\sin \theta \right),\theta \in \left[ 0,2\pi \right] $ .
The center of the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ is at $ (0,\ 0) $ .

Use the formula for the distance between a point and a line, and find its maxima/minima using derivatives.

Complete step by step solution:
Let's say that $ \left( a\cos \theta ,\ b\sin \theta \right),\theta \in \left[ 0,2\pi \right] $ is a point on the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ .

The equation of the normal through this point will be:
$ \dfrac{{{a}^{2}}x}{a\cos \theta }-\dfrac{{{b}^{2}}y}{b\sin \theta }={{a}^{2}}-{{b}^{2}} $
⇒ $ ax\sec \theta -by\csc \theta ={{a}^{2}}-{{b}^{2}} $

Comparing this with the general equation of a line $ Ax+By+C=0 $ , we have:
$ A=a\sec \theta ,\ B=-b\csc \theta ,\ C=-\left( {{a}^{2}}-{{b}^{2}} \right) $

The distance between the center $ (0,0) $ and the above line will be:
$ d=\left| \dfrac{A(0)+B(0)+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| $
⇒ $ d=\left| \dfrac{-\left( {{a}^{2}}-{{b}^{2}} \right)}{\sqrt{{{(a\sec \theta )}^{2}}+{{(b\csc \theta )}^{2}}}}
\right| $
⇒ $ d=\left| \dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\csc }^{2}}\theta }}
\right| $
Since a and b are constants, the above distance will be maximum/minimum when the denominator is minimum/maximum.
Differentiating the term $ {{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\csc }^{2}}\theta $ w.r.t. $ \theta $ and equating to 0, we get:
$ {{a}^{2}}(2\sec \theta )(\sec \theta \tan \theta )+{{b}^{2}}(2\csc \theta )(-\csc \theta \cot \theta )=0 $
⇒ $ {{a}^{2}}{{\sec }^{2}}\theta \tan \theta -{{b}^{2}}{{\csc }^{2}}\theta \cot \theta =0 $

Writing $ \sec \theta =\dfrac{1}{\cos \theta } $ and $ \csc \theta =\dfrac{1}{\sin \theta } $ , we get:
⇒ $ \dfrac{{{a}^{2}}}{{{b}^{2}}}=\left( \dfrac{1}{{{\sin }^{2}}\theta }\cot \theta \right)\left( {{\cos
}^{2}}\theta \cot \theta \right) $
⇒ $ \dfrac{{{a}^{2}}}{{{b}^{2}}}={{\cot }^{4}}\theta $
⇒ $ \dfrac{a}{b}={{\cot }^{2}}\theta $

We can also see that the derivative of ${{a}^{2}}{{\sec }^{2}}\theta \tan \theta -{{b}^{2}}{{\csc
}^{2}}\theta \cot \theta $ is ${{a}^{2}}{{\sec }^{2}}x\left( 2{{\tan }^{2}}x+{{\sec }^{2}}x
\right)+{{b}^{2}}{{\csc }^{2}}x\left( {{\csc }^{2}}x+2{{b}^{2}}{{\cot }^{2}}x \right)$ which is always positive.

Therefore, the minimum value of the expression occurs at $\dfrac{a}{b}={{\cot }^{2}}\theta $ .
And the distance is $ d=\left| \dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\csc
}^{2}}\theta }} \right| $
Using the identities $ 1+{{\tan }^{2}}\theta =se{{c}^{2}}\theta $ and $ 1+{{\cot }^{2}}\theta ={{\csc
}^{2}}\theta $ , we get.
⇒ $ d=\left| \dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}\left( 1+{{\tan }^{2}}\theta \right)+{{b}^{2}}\left(
1+{{\cot }^{2}}\theta \right)}} \right| $
Substituting $ \dfrac{a}{b}={{\cot }^{2}}\theta $ and simplifying further, we get:
⇒ $ d\le \left| \dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}\left( 1+\tfrac{b}{a} \right)+{{b}^{2}}\left(
1+\tfrac{a}{b} \right)}} \right| $
⇒ $ d\le \left| \dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}+2ab+{{b}^{2}}}} \right| $
⇒ $ d\le \left| \dfrac{(a-b)(a+b)}{a+b} \right| $
⇒ $ d\le \left| a-b \right| $ , hence proved.

Note:The standard form, $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , of an ellipse in Cartesian coordinates assumes that the origin is the center of the ellipse, the x-axis is the major axis, the foci are the points $ {{F}_{1}}=(c,\ 0),\ {{F}_{2}}=(-c,\ 0) $ and the vertices are $ {{V}_{1}}=(a,\ 0),\ {{V}_{2}}=(-a,\
0) $ .
The eccentricity of the ellipse is $ e=\dfrac{c}{a}=\sqrt{1-{{\left( \dfrac{b}{a} \right)}^{2}}} $ , for $ a>b $ . Tangent to an Ellipse: The equation of the tangent to the ellipse $
\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , at a point $ ({{x}_{1}},{{y}_{1}}) $ , is given by:
$ \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1 $ .
If the line $ y=mx+c $ touches the ellipse, then $ {{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}} $ .

The distance between a point $ P({{x}_{1}},{{y}_{1}}) $ and the line $ ax+by+c=0 $ is given by:
$ Distance=\dfrac{|a{{x}_{1}}+b{{y}_{1}}+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}} $