Question

The normal at

P

meets the axes in

G

and

g;

show that the loci of the middle point of

P G

and

G g

are respectively the ellipses

\frac{4 x^{2}}{a^{2} {(1 + e^{2})}^{2}} + \frac{4 y^{2}}{b^{2}} = 1

and

a^{2} x^{2} + b^{2} y^{2} = \frac{1}{4} {(a^{2} - b^{2})}^{2} .

Accepted Answer

The equation to normal at any point $P = (a \cos ϕ, b \sin ϕ)$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is
$a x \sec ϕ - b y cosec ϕ = a^{2} - b^{2} . \dots (1)$
Let it meet the axes at $G$ and $g$ respectively.
Solving equation $(1)$ with x-axis, i.e., $y = 0,$ we get $x = \frac{a^{2} - b^{2}}{a} \cos ϕ$
So, the co-ordinates of $G$ are $(\frac{a^{2} - b^{2}}{a} \cos ϕ, 0)$
Solving equation $(1)$ with y-axis, i.e., $x = 0,$ we get the co-ordinates of $g$ as $(0, \frac{a^{2} - b^{2}}{- b} \sin ϕ)$
If the co-ordinates of the middle point of $P G$ be $(x, y)$ , then
$x = \frac{1}{2} [a \cos ϕ + \frac{a^{2} - b^{2}}{a} \cos ϕ]$ and $y = \frac{b s i n ϕ}{2}$
Hence, $\cos ϕ = \frac{2 x a}{2 a^{2} - b^{2}} = \frac{2 x a}{2 a^{2} - a^{2} (1 - e^{2})} = \frac{2 x a}{a^{2} + a^{2} e^{2}} = \frac{2 x}{a (1 + e^{2})}$
And $\sin ϕ = \frac{2 y}{b}$
Now squaring and adding, we get
$\frac{4 x^{2}}{a^{2} {(1 + e^{2})}^{2}} + \frac{4 y^{2}}{b^{2}} = \cos^{2} ϕ + \sin^{2} ϕ$
$\Rightarrow$ $\frac{4 x^{2}}{a^{2} {(1 + e^{2})}^{2}} + \frac{4 y^{2}}{b^{2}} = 1$

The equation to normal at any point $P = (a \cos ϕ, b \sin ϕ)$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is
$a x \sec ϕ - b y cosec ϕ = a^{2} - b^{2} . \dots (1)$
Let it meet the axes at $G$ and $g$ respectively.
Solving equation $(1)$ with x-axis, i.e., $y = 0,$ we get $x = \frac{a^{2} - b^{2}}{a} \cos ϕ$
So, the co-ordinates of $G$ are $(\frac{a^{2} - b^{2}}{a} \cos ϕ, 0)$
Solving equation $(1)$ with y-axis, i.e., $x = 0,$ we get the co-ordinates of $g$ as $(0, \frac{a^{2} - b^{2}}{- b} \sin ϕ)$
If the co-ordinates of the middle point of $G g$ be $(x, y),$ then
$x = \frac{a^{2} - b^{2}}{2 a} \cos ϕ$ and $y = \frac{a^{2} - b^{2}}{- 2 b} \sin ϕ .$
Hence, $\cos ϕ = \frac{2 a x}{a^{2} - b^{2}}$ and $\sin ϕ = \frac{- 2 b y}{a^{2} - b^{2}}$
Squaring and adding, we get
$\frac{4 a^{2} x^{2}}{{(a^{2} - b^{2})}^{2}} + \frac{4 b^{2} y^{2}}{{(a^{2} - b^{2})}^{2}} = \cos^{2} ϕ + \sin^{2} ϕ = 1$
$\Rightarrow a^{2} x^{2} + b^{2} y^{2} = \frac{1}{4} {(a^{2} - b^{2})}^{2} .$

EXAMPLES XXXIII

S L Loney Mathematics Solutions for Exercise - EXAMPLES XXXIII

Questions from EXAMPLES XXXIII with Hints & Solutions

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