Find an equation of a parabola with a vertex at the origin and directrix X = 2.5
Mathematics
nancyvalencia36
Question
Find an equation of a parabola with a vertex at the origin and directrix X = 2.5
2 Answer

1. User Answers surjithayer
vertex=(0,0)
directrix is x=2.5
or x2.5=0
so focus S is (2.5,0)
let P(x,y)be any point on the parabola.
Let M be the foot of perpendicular from P(x,y) on the directrix.
then SP=PM
[tex] \sqrt{(x(2.5))^2+(y0)^2} = \frac{x2.5}{( \sqrt{1})^2 }  squaring both sides[/tex][tex](x+2.5)^2+y^2=(x2.5)^2 ,
x^2+6.25+5x+y^2=x^2+6.255x [/tex]
[tex]y^2=x^2+6.255xx^26.255x=10x so eq. is y^{2} =10x[/tex] 
2. User Answers lhabdulsamirahmed
To solve this problem, we just need to substitute the values into the equation and solve. The equation of the parabola is equal to y^2 = 10x
Equation of a Parabola
The standard equation of a parabola is given as
[tex](xh)^2 = 4a(y  k)\\[/tex]
But in this question, the vertex is starting from the origin and the directrix is at 2.5
 vertex = (0,0)
 directrix x = 2.5
 h = 0
 k = 0
 a = 2.5
substituting the values into the equation;
[tex]y^2 = 4*(2.5)x\\ \\y^2 = 10x\\x = \frac{1}{10}y^2[/tex]
The equation of the parabola is equal to y^2 = 10x
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