What is the atomic radius of a BCC crystal structure?
2 answers:
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Answer:
Step-by-step explanation:
∵ The quadratic equation form is :
y = [1/2(b - k)] (x - a)² + (b + k)/2
Where (a , b) is the focus and directrix y = k
∵ The focus is (4 , -3) and directix is y = -6
∵
∴
∴
another way:
Assume that (x , y) is the general point on the parabola
∵ The distance between the directrix and (x , y) = the distance between the focus and (x , y)
By using the distance rule:
∵ (y - -6)² = (x - 4)² + (y - -3)² ⇒ (y + 6)² = (x - 4)² + (y + 3)
∴ y² + 12y + 36 = (x - 4)² + y² + 6y + 9
∴ 12y - 6y = (x - 4)² + 9 - 36
∴ 6y = (x - 4)² - 27 ⇒ ÷ 6
∴ y = 1/6 (x - 4)² - 9/2
Answer:
B= 4.6cm
Step-by-step explanation:
A=