l/r=1+ecosθ eching

$$\frac { l } { r } = 1 + e \cos \theta$$

l uchun yechish

$$l=r\left(e\cos(\theta )+1\right)$$

$$r\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\frac{1}{r}l=e\cos(\theta )+1$$

Ikki tarafini r^{-1} ga bo‘ling.

$$\frac{\frac{1}{r}lr}{1}=\frac{\left(e\cos(\theta )+1\right)r}{1}$$

r^{-1} ga bo'lish r^{-1} ga ko'paytirishni bekor qiladi.

$$l=\frac{\left(e\cos(\theta )+1\right)r}{1}$$

1+e\cos(\theta ) ni r^{-1} ga bo'lish.

$$l=r\left(e\cos(\theta )+1\right)$$

r uchun yechish

$$\left\{\begin{matrix}r=\frac{l}{e\cos(\theta )+1}\text{, }&l\neq 0\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }\theta =2\pi n_{2}+\arccos(\frac{1}{e})+\pi \text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}-\arccos(\frac{1}{e})+\pi \\r\neq 0\text{, }&\left(\exists n_{4}\in \mathrm{Z}\text{ : }\theta =2\pi n_{4}+\arccos(\frac{1}{e})+\pi \text{ or }\exists n_{3}\in \mathrm{Z}\text{ : }\theta =2\pi n_{3}-\arccos(\frac{1}{e})+\pi \right)\text{ and }l=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

r qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini r ga ko'paytirish.

$$l=r+e\cos(\theta )r$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$r+e\cos(\theta )r=l$$

r'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1+e\cos(\theta )\right)r=l$$

Tenglama standart shaklda.

$$\left(e\cos(\theta )+1\right)r=l$$

Ikki tarafini 1+e\cos(\theta ) ga bo‘ling.

$$\frac{\left(e\cos(\theta )+1\right)r}{e\cos(\theta )+1}=\frac{l}{e\cos(\theta )+1}$$

1+e\cos(\theta ) ga bo'lish 1+e\cos(\theta ) ga ko'paytirishni bekor qiladi.

$$r=\frac{l}{e\cos(\theta )+1}$$

r qiymati 0 teng bo‘lmaydi.

$$r=\frac{l}{e\cos(\theta )+1}\text{, }r\neq 0$$