Question

In an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

, show that the perpendiculars from the center upon all the chords which join the ends of the perpendicular diameters, are of constant length.

Accepted Answer

Let the ellipse be given as,

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

And, let the perpendicular semi-diameters be the lines $O L$ and $O M$ .

Let the equation of the line $L M$ be given as,

$x \cos α + y \sin α = p$ (where the length of the perpendicular from the centre $O$ upon the line is $p$ )

Homogenising the equation of the ellipse with the help of the line $L M$ , we get the combined equation of the lines $O L$ and $O M$ as,

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = {(\frac{x \cos α + y \sin α}{p})}^{2}$

$x^{2} (\frac{1}{a^{2}} - \frac{\cos^{2} α}{p^{2}}) - \frac{2 x y \sin α \cos α}{p^{2}} + y^{2} (\frac{1}{b^{2}} - \frac{\sin^{2} α}{p^{2}}) = 0$

The lines $O L$ and $O M$ are mutually perpendicular, if the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero.

$\Rightarrow \frac{1}{a^{2}} - \frac{\cos^{2} α}{p^{2}} + \frac{1}{b^{2}} - \frac{\sin^{2} α}{p^{2}} = 0$

$\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{\sin^{2} α + \cos^{2} α}{p^{2}}$

$\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}$

$\Rightarrow p = \frac{a b}{\sqrt{a^{2} + b^{2}}}$

Hence, the perpendiculars from the center upon all the chords, which join the ends of the perpendicular diameters, are of constant length.

S L Loney Solutions for Chapter: The Ellipse, Exercise 1: EXAMPLES XXXII

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Ellipse, Exercise 1: EXAMPLES XXXII

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