Mathematics

Question

Find an equation of a parabola with a vertex at the origin and directrix X = 2.5

2 Answer

  • vertex=(0,0)
    directrix is x=2.5
    or x-2.5=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+(y-0)^2} =| \frac{x-2.5}{( \sqrt{1})^2 } | squaring both sides[/tex][tex](x+2.5)^2+y^2=(x-2.5)^2 ,
    x^2+6.25+5x+y^2=x^2+6.25-5x [/tex]
    [tex]y^2=x^2+6.25-5x-x^2-6.25-5x=-10x so eq. is y^{2} =-10x[/tex]


  • 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](x-h)^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

    Learn more on equation of a parabola here;

    https://brainly.com/question/17987697

    #SPJ2

NEWS TODAY

You May Be Interested