Asosiy tarkibga o'tish

4.3-bo'lim : Ellipslar

2. Quyidagi ellipsning grafigini chizing.

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

Bu erda birinchi qadam bizning tenglamamizni ellipsning standart shakli bilan solishtirish va barcha muhim ma'lumotlarni aniqlashdir. Malumot uchun bu erda ellipsning standart shakli keltirilgan.

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

Taqqoslashda yordam berish uchun keling, tenglamamiz standart shaklga o'xshash bo'lishi uchun uni biroz qayta yozamiz.

\[\frac{{{{\left( {x - 0} \right)}^2}}}{1} + \frac{{{{\left( {y - 1} \right)}^2}}}{4} = 1\]

Bizga maxrajdagi "1" kerak emas va birinchi hadning hisobidagi "-0" kerak emas, lekin bu taqqoslash jarayonida yordam berishi mumkin.

Bizning tenglamamizni bu bilan solishtirsak, biz quyidagi ma'lumotlarga ega ekanligimizni ko'rishimiz mumkin.

\[h = 0\hspace{0.25in}k = 1\hspace{0.25in}a = 1\hspace{0.25in}b = 2\]
2-qadam

Birinchi bosqichda topilgan ma'lumotlar bilan biz ellipsning markazi \(\left( {0,1} \right)\) ekanligini ko'rishimiz mumkin.

Eng o'ngda, eng chapda, eng yuqorida va pastki qismida eng ko'p nuqtalar, keyin,

\[\begin{align*} & {\mbox{Right Most Point :}}\,\,\left( {1,1} \right)\\ & {\mbox{Left Most Point :}}\,\,\left( { - 1,1} \right)\\ & {\mbox{Top Most Point :}}\,\,\left( {0,3} \right)\\ & {\mbox{Bottom Most Point :}}\,\,\left( {0, - 1} \right)\end{align*}\]
3-qadam

Bu erda biz yuqorida topilgan barcha nuqtalarni o'z ichiga olgan ellipsning eskizi.

The graph of the ellipse with center \(\left( {0,1} \right)\).  The center is marked with a dot and the four points found above are marked and labeled on the ellipse.  They are the right most point \(\left( {1,1} \right)\), the left most point \(\left( { - 1,1} \right)\), the top most point \(\left( {0,3} \right)\) and the bottom most point \(\left( {0, - 1} \right)\).  Because the ellipse reaches further up and down than it does left and right it is taller than it is wide.