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Question Number 9199 by tawakalitu last updated on 22/Nov/16

If  r^2  = (x + ea)^2  + y^2  and s^2  = (x − ea)^2  + y^2   and r + s = 2a, Prove that:  r = a + ex,  s = a − ex, and that,  x^2 (1 − e^2 ) + y^2  = a^2 (1 − e^2 )

$$\mathrm{If}\:\:\mathrm{r}^{\mathrm{2}} \:=\:\left(\mathrm{x}\:+\:\mathrm{ea}\right)^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{s}^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\mathrm{ea}\right)^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{r}\:+\:\mathrm{s}\:=\:\mathrm{2a},\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{r}\:=\:\mathrm{a}\:+\:\mathrm{ex},\:\:\mathrm{s}\:=\:\mathrm{a}\:−\:\mathrm{ex},\:\mathrm{and}\:\mathrm{that}, \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}\:−\:\mathrm{e}^{\mathrm{2}} \right)\:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}\:−\:\mathrm{e}^{\mathrm{2}} \right) \\ $$

Commented by 123456 last updated on 22/Nov/16

r^2 +s^2 =2y^2 +(x+ea)^2 +(x−ea)^2   r^2 +s^2 =2y^2 +x^2 +2ea+(ea)^2 +x^2 −2ea+(ea)^2   r^2 +s^2 =2[x^2 +y^2 +(ea)^2 ]  r+s=2a  (r+s)^2 =4a^2   r^2 +s^2 +2rs=4a^2   2[x^2 +y^2 +(ea)^2 +rs]=4a^2   x^2 +y^2 +(ea)^2 +rs=2a^2

$${r}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} +\left({x}+{ea}\right)^{\mathrm{2}} +\left({x}−{ea}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{ea}+\left({ea}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{ea}+\left({ea}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{2}\left[{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({ea}\right)^{\mathrm{2}} \right] \\ $$$${r}+{s}=\mathrm{2}{a} \\ $$$$\left({r}+{s}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +{s}^{\mathrm{2}} +\mathrm{2}{rs}=\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\mathrm{2}\left[{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({ea}\right)^{\mathrm{2}} +{rs}\right]=\mathrm{4}{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({ea}\right)^{\mathrm{2}} +{rs}=\mathrm{2}{a}^{\mathrm{2}} \\ $$

Commented by tawakalitu last updated on 23/Nov/16

I appreciate sir. God bless you.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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