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数学 高校生

レムニスケートについての問題です。蛍光ペンを引いたところが分かりません。なぜf(x,y)=0となるのでしょうか。

直交座標のまま対称性を調べ,その結果 0<0< の範囲で概形を調べる。 OO0 重要例題72 レムニスケートの極方程式 曲線(x+y)?=x-y について,次の問いに答えよ。 (1) 与えられた曲線がx軸, y軸,原点に関して刈杯であることあ、 116 (2) 与えられた曲線の極方程式を求め, 概形をかけ。 CHARTOSOLUTION 座標の選定 → 極座標 対称性 → 直交座標,概形 - (1) f(x, y)=(x°+y)?-(x°-y°) とすると, 与えられた曲線 の方程式は (解答 の f(x, y)=0 (x, -y)=(-x, )=f(-x, -y)=f(x, y) であるから 曲線のは,x軸,y軸, 原点に関してそれぞれ対称である。 (2) 与式にx=rcos e, y=rsin6, x°+y?=r? を代入すると ()=(cos'0-sin°0) PC ゆえに (r2-cos 20)=0 I Cos'0-sin'9=as IQ よって ア=0 または r=cos 20 IS ア=0 は=cos20 に含まれるから,求める極形式は r=cos20 曲線のの対称性から, r20, 0S0sの範囲で考える。 x20, y20 の疑 COS える。 また,パ20 から Cos 2020 ゆえに,曲線の存在範囲は 0S0S- 0| 0 Tπ T||T 12 8 6|4 ロ /3 0= 121 0 2 0=。 1 2 2 これらをもとにして,第1象限にお ける曲線のをかき,それとx軸,y 軸,原点に関して対称な曲線もかき 加えると,曲線の概形は右の図のようになる。 1x 0=0 0=2 linf. この曲線を、 レムニスケートと PF PRACTICO たま |o う 。

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