From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
Important notes:
3 sides 1 angle - COSINE RULE
2 sides 2 angle - SINE RULE
since, the question wants to find the length of BC. In the end we will have 3 sides and 1 angle and use cosine rule
formula of cosine rule:
a² = b² + c² - 2bc Cos A° (to find the length)
Cos A° = b² + c² - a² / 2bc ( to find the angle, if there is given three sides and have to find the angle)
So just substitute,
a² = 13² + 15² - 2(13)(15) Cos 95°
a = 20.6 or 21
Let
The origin of coordinates the tree
r1 = vector position of the child 1.
r2 = vector position of the child 2
Child 1:
r1 = (12i + 12j)
Child 2:
r2 = (-18i + 11j)
The scalar product will be given by:
r1.r2 = ((12) * (- 18)) + ((12) * (11)) = - 84
The scalar product of their net displacements from the tree is -84m ^ 2
Answer:
Should be B
Step-by-step explanation: