Asosiy tarkibga o'tish

4.4-bo'lim : Giperbolalar

1. Quyidagi giperbolaning grafigini chizing.

\[\frac{{{y^2}}}{{16}} - \frac{{{{\left( {x - 2} \right)}^2}}}{9} = 1\]
Yechim

Bu erda birinchi qadam bizning tenglamamizni giperbolaning standart shakli bilan solishtirish va barcha muhim ma'lumotlarni aniqlashdir. Malumot uchun bu erda bizda mavjud bo'lgan giperbolaning standart shakli keltirilgan.

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

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

\[h = 2\hspace{0.25in}k = 0\hspace{0.25in}a = 3\hspace{0.25in}b = 4\]

\(y\) atamasi ijobiy atama bo'lgani uchun biz bilamizki, bu giperbola yuqoriga va pastga ochiladi.

2-qadam

Birinchi bosqichda topilgan ma'lumotlar bilan biz giperbolaning markazi \(\left( {2,0} \right)\) ekanligini ko'rishimiz mumkin.

Giperbolaning uchlari: \(\left( {2, - 4} \right)\) va \(\left( {2,4} \right)\).

Ikki asimptotaning tenglamalari:

\[y = 0 + \frac{4}{3}\left( {x - 2} \right) = \frac{4}{3}x - \frac{8}{3}\hspace{0.25in}y = 0 - \frac{4}{3}\left( {x - 2} \right) = - \frac{4}{3}x + \frac{8}{3}\]
3-qadam

Mana, yuqorida topilgan nuqtalar va asimptotalarni o'z ichiga olgan giperbolaning eskizi.

The graph of the hyperbola \(\frac{{{y^2}}}{{16}} - \frac{{{{\left( {x - 2} \right)}^2}}}{9} = 1\).  It is centered at \(\left( {2,0} \right)\) and opens up and down with vertices at \(\left( {2,4} \right)\) and \(\left( {2, - 4} \right)\), both of which are marked and labeled.  The two blue dashed asymptotes cross at the center and are labeled \(y = \frac{4}{3}x - \frac{8}{3}\) and \(y = - \frac{4}{3}x + \frac{8}{3}\), and the two branches of the hyperbola get closer and closer to them as you move away from the center.