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4.3-bo'lim: Ellipslar

Avvalgisida bo'limi biz aylanalarning grafikasini ko'rib chiqdik va doiralar haqiqatan ham ellipslarning alohida holatlari bo'lganligi sababli, bizda ellipslarning grafigini yaratish uchun ko'plab vositalar mavjud. Bizni boshlash uchun bu erda kerak bo'lgan narsa - bu standart shakl ellips va uni qanday talqin qilish haqida ozgina ma'lumot.

Mana ellipsning standart shakli.

\[\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\]

E'tibor bering, standart shaklda bo'lish uchun o'ng tomon 1 bo'lishi kerak. \(\left( {h,k} \right)\) nuqtasi deyiladi markaz ellipsdan.

Ellipsning grafigini tuzish uchun bizga eng o'ng, eng chap, eng yuqori va pastki eng ko'p nuqtalar kerak bo'ladi. Ularga ega bo'lgach, ellipsda eskiz chizishimiz mumkin. Mana bu nuqtalarni topish uchun formulalar.

\[\begin{align*} & {\mbox{right most point : }}\left( {h + a,k} \right)\\ & {\mbox{left most point : }}\left( {h - a,k} \right)\\ & {\mbox{top most point : }}\left( {h,k + b} \right)\\ & {\mbox{bottom most point : }}\left( {h,k - b} \right)\end{align*}\]

Shuni e'tiborga oling a ostidagi sonning kvadrat ildizidir x muddat va biz markazdan o'ngga va chapga siljitadigan miqdor. Shuningdek, b ostidagi sonning kvadrat ildizidir y muddat va biz markazdan yuqoriga yoki pastga siljiydigan miqdor.

Keling, bir nechta grafiklarni chizamiz.

1-misol Quyidagi ellipslarning har birining grafigini chizing.
  1. \(\displaystyle \frac{{{{\left( {x + 2} \right)}^2}}}{9} + \frac{{{{\left( {y - 4} \right)}^2}}}{{25}} = 1\)
  2. \(\displaystyle \frac{{{x^2}}}{{49}} + \frac{{{{\left( {y - 3} \right)}^2}}}{4} = 1\)
  3. \(4{\left( {x + 1} \right)^2} + {\left( {y + 3} \right)^2} = 1\)
a \(\displaystyle \frac{{{{\left( {x + 2} \right)}^2}}}{9} + \frac{{{{\left( {y - 4} \right)}^2}}}{{25}} = 1\) Yechim

Shunday qilib, bu ellipsning markazi \(\left( { - 2,4} \right)\) va odatdagidek bu erda belgilar bilan ehtiyot bo'ling! Shuningdek, bizda \(a = 3\) va \(b = 5\) mavjud. Demak, nuqtalar,

\[\begin{align*} & {\mbox{right most point : }}\left( {1,4} \right)\\ & {\mbox{left most point : }}\left( { - 5,4} \right)\\ & {\mbox{top most point : }}\left( { - 2,9} \right)\\ & {\mbox{bottom most point : }}\left( { - 2, - 1} \right)\end{align*}\]

Mana bu ellipsning eskizi.

The domain of this graph is from -5 to 1 while the range is from -1 to 9.  This graph looks like a circle that has been stretched vertically.  The center is at (-2,4) and the top/bottom points are (-2,9)/(-2,-1) while the right/left points are (1,4)/(-5,4).

b \(\displaystyle \frac{{{x^2}}}{{49}} + \frac{{{{\left( {y - 3} \right)}^2}}}{4} = 1\) Yechim

Ushbu qismning markazi \(\left( {0,3} \right)\) va bizda \(a = 7\) va \(b = 2\) mavjud. Bizga kerak bo'lgan nuqtalar,

\[\begin{align*}& {\mbox{right most point : }}\left( {7,3} \right)\\ & {\mbox{left most point : }}\left( { - 7,3} \right)\\ & {\mbox{top most point : }}\left( {0,5} \right)\\ & {\mbox{bottom most point : }}\left( {0,1} \right)\end{align*}\]

Mana bu ellipsning eskizi.

The domain of this graph is from -5 to 1 while the range is from -1 to 9.  This graph looks like a circle that has been stretched horizontally.  The center is at (0,3) and the top/bottom points are (0,1)/(0,5) while the right/left points are (7,3)/(7,-3).

c \(4{\left( {x + 1} \right)^2} + {\left( {y + 3} \right)^2} = 1\) Yechim

Endi bu ellips bilan biz biroz ehtiyot bo'lishimiz kerak, chunki u hali standart shaklda emas. Mana bu ellips uchun standart shakl.

\[\frac{{{{\left( {x + 1} \right)}^2}}}{{\frac{1}{4}}} + {\left( {y + 3} \right)^2} = 1\]

E'tibor bering, birinchi hadning hisoblagichida 4 koeffitsientini olish uchun maxrajda \(\frac{1}{4}\) bo'lishi kerak. Shuni ham yodda tutingki, bizda hatto kasr ham yo'q y muddat. Bu maxrajda aslida 1 borligini bildiradi. Agar bu erda nima bo'layotganini ko'rish foydali bo'lsa, biz buni qo'yishimiz mumkin.

\[\frac{{{{\left( {x + 1} \right)}^2}}}{{\frac{1}{4}}} + \frac{{{{\left( {y + 3} \right)}^2}}}{1} = 1\]

Shunday qilib, bu shaklda biz markaz \(\left( { - 1, - 3} \right)\) va \(a = \frac{1}{2}\) va \(b = 1\) ekanligini ko'rishimiz mumkin. Ushbu ellips uchun nuqtalar,

\[\begin{align*} & {\mbox{right most point : }}\left( { - \frac{1}{2}, - 3} \right)\\ & {\mbox{left most point : }}\left( { - \frac{3}{2}, - 3} \right)\\ & {\mbox{top most point : }}\left( { - 1, - 2} \right)\\ & {\mbox{bottom most point : }}\left( { - 1, - 4} \right)\end{align*}\]

Mana bu ellips.

The domain of this graph is from -1.5 to 0.5 while the range is from -4 to 1.  This graph looks like a circle that has been stretched vertically.  The center is at (-1,-3) and the top/bottom points are (-1,-2)/(-1,-4) while the right/left points are (-1/2,-3)/(-3/2,-3).

Va nihoyat, keling, ushbu bo'limning boshida qilingan sharhga murojaat qilaylik. Biz aylanalar ellipsning alohida holatidan boshqa narsa emasligini aytdik. Buni ko'rish uchun \(a = b\) deb faraz qilaylik. Bu holatda bizda,

\[\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1\]

E'tibor bering, biz \(a = b\)-ni tan oldik va foydalandik a ikkala holatda ham. Endi biz maxrajlarni aniqlaymiz,

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}\]

Bu markazi \(\left( {h,k} \right)\) va radiusli aylananing standart shakli a. Shunday qilib, doiralar haqiqatan ham ellipslarning alohida holatlaridir.