Deriving the equation of the parabloid formula
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Given a parabola with focal length f, we can derive the equation of the parabola. (see figure on right). We assume the origin (0,0) of the coordinate system is at the parabola's vertex.
For any point (x,y) on the parabola, the two blue lines labelled d have the same length, because this is the definition of a parabola. So we can find an equation for each of them, set them equal to each other and simplify to find the parabola's equation.
The top line d is the hypotenuse of the small right triangle. The horizontal side of the triangle has a length of x. The vertical side has a length of (y–f).
From the Pythagorean theorem:
d=√x2(y−f2
Turning to the vertical line d. From Parabola definition (focus-directrix) we know the vertex is always half way between the focus and the directrix, so:
d=y+f
So we now have two equations for d. We know they are equal, so we can set the equations equal to each other:
y+f=√x2+(y−f 2)
We now simplify this, aiming to get y on the left side. So, first square both sides to get rid of theradical:
(y+f)2=2+(y−f2
Expanding the squared expressions
y2+f+f2x +22fyf
The y2 and f2 terms on each side cancel:
2fy=x2−2fy
Add 2fy to each side
4fy=x2
Divide both sides by 4f
y=14fx2
Hope this question's answer are helpful
and press the point of thanks
And please bainalest
For any point (x,y) on the parabola, the two blue lines labelled d have the same length, because this is the definition of a parabola. So we can find an equation for each of them, set them equal to each other and simplify to find the parabola's equation.
The top line d is the hypotenuse of the small right triangle. The horizontal side of the triangle has a length of x. The vertical side has a length of (y–f).
From the Pythagorean theorem:
d=√x2(y−f2
Turning to the vertical line d. From Parabola definition (focus-directrix) we know the vertex is always half way between the focus and the directrix, so:
d=y+f
So we now have two equations for d. We know they are equal, so we can set the equations equal to each other:
y+f=√x2+(y−f 2)
We now simplify this, aiming to get y on the left side. So, first square both sides to get rid of theradical:
(y+f)2=2+(y−f2
Expanding the squared expressions
y2+f+f2x +22fyf
The y2 and f2 terms on each side cancel:
2fy=x2−2fy
Add 2fy to each side
4fy=x2
Divide both sides by 4f
y=14fx2
Hope this question's answer are helpful
and press the point of thanks
And please bainalest
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