Asosiy tarkibga o'tish

12.4-bo'lim : To'rtburchak yuzalar

4. Quyidagi to'rtburchak sirtni chizing.

\[{y^2} = 4{x^2} + 16{z^2}\]
Yechim

Bu konusning markazi \(y\) o'qi bo'lib, \(x\) va \(z\) atamalarining koeffitsientlari har xil bo'lgani uchun sirtning kesmalari ellips bo'ladi.

Eslatmalarda berilgan tenglamalarni boshqa koordinata o'qlariga "tarjima" qilishingiz mumkinligiga ishonch hosil qiling. Koordinatalar o'qlaridan birida markazlashganda ular qanday ko'rinishini bilganingizdan so'ng, oddiy va bashorat qilinadigan o'zgaruvchan o'zgarishlar ularni boshqa koordinata o'qlarida markazlashtiradi.

Bu erda mintaqaning bir nechta eskizlari. Biz ularga an'anaviy o'qlarni, shuningdek, sirtni tasavvur qilish uchun "qutili" boltalarni berdik.

A three dimensional sketch of the surface \({y^2} = 4{x^2} + 16{z^2}\) drawn with the traditional axes. It is an elliptic cone centered on the y-axis with its point at the origin, opening out in both the positive and negative y directions.
A three dimensional sketch of the same elliptic cone \({y^2} = 4{x^2} + 16{z^2}\), this time drawn with “boxed” axes to make the surface easier to visualize. It is centered on the y-axis with its point at the origin and opens out in both directions along that axis.